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Crossing Over of John Edward’s Chromosomes
Linkage & Mapping
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
Linkage
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
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
Figure 5.1
Linkage Prevents Independent Assortment
Figure 5.1
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
Crossing Over May Produce Recombinant
Phenotypes
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
Purple flowers, long pollenPurple flowers, round pollenRed flowers, long pollenRed flowers, round pollen
296192785
15.61.01.44.5
240808027
9331
Observednumber
ObservedRatio
ExpectedRatio
Expectednumber
Purple flowers,long pollen (PPLL)
F2 offspring phenotypes
F1 offspring
Purple flowers,long pollen (PpLl)
Self-fertilization
x
Red flowers,round pollen (ppll)
Example of Linkage
(9:3:3:1)P
P
NPNP
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
Linkage to a Particular
Chromosome
Tan body, red eyes, normal wingsTan body, red eyes, miniature wingsTan body, white eyes, normal wingsTan body, white eyes, miniature wingsYellow body, red eyes, normal wingsYellow body, red eyes, miniature wingsYellow body, white eyes, normal wingsYellow body, white eyes, miniature wings
439208
1570
178365
319193
01150
139335
758401
116120
317700
Females Males Total
y w m/y w m
y+w+ m+/ y w m
F2 generation
F1 generation
F1 generation contains wild-typefemales and yellow-bodied,white-eyed, miniature-wingedmales.
x
y w m / Y
x
y+ w+ m+ Y
Sex linkage of all traits places them all on X
chromosome
/
P
P
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Morgan observed a much higher proportion of the combinations of traits found in the parental generation
P Males
P Females
Morgan’s explanation: All three genes are located on the X chromosome Therefore, they tend to be transmitted together as a unit
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?
Morgan Provided Evidence for the Linkage of Several X-linked Genes
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
Reorganize Morgan’s data considering pairs of genes separately
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
But this nonparental combination was rare
Figure 5.4
These parental phenotypes are the most common offspring
because the genes are far apart
These recombinant offspring are not uncommon
Figure 5.4
because the genes are very close together
These recombinant offspring are fairly uncommon
These recombinant offspring are very unlikely1 out of 2,205
it is product of single cross over probabilities
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
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.”
Alfred Sturtevant’s Analysis
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 offspringX 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
Linkage and Genetic Maps
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 e = ebony body color + = gray body color
s = short bristles + = normal bristles
One parent double recessive (homozygous recessive at both loci) – s/s ; e/e
Other parent is heterozygous at both loci (+/s ; +/e)
Linkage Analysis and Mapping
5-47Figure 5.9
Chromosomes are the product of a crossover during
meiosis in the heterozygous parent
Recombinant offspring are fewer
in number than nonrecombinant
offspring
The phenotype data are used to estimate the distance between the two loci
Map distance =
Number of recombinant offspring
Total number of offspringX 100
Therefore, the s and e genes are 12.3 map units apart from each other along the same chromosome
76 + 75
542 + 537 + 76 + 75X 100=
= 12.3 map units
Linkage Analysis and Mapping
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
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Trihybrid Crosses
b = black body color b+ = grey body color
pr = purple eye color pr+ = red eye color
vg = vestigial wings vg+ = normal wings
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Step 1: Cross two true-breeding strains that differ at three loci.
Female is mutant for all three traits
Male is homozygous wildtype for all three
traits
The goal in this step is to obtain F1 individuals that are heterozygous for all three genes
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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
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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
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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
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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
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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
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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
30 + 28
1,005= 0.058
Single crossover between b and pr
61 + 60
1,005= 0.120
Single crossover between pr and vg
1 + 2
1,005= 0.003
Double crossover, between b and pr, and between pr and vg
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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 = (0.058 + 0.003) X 100 = 6.1 mu
To calculate the distance between pr and vg Map distance = (0.120 + 0.003) 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 = (0.058 + 0.120 + 2[0.003]) X 100
= 18.4 mu
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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
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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 = 6.1 + 12.3 = 18.4 mu
The product rule allows us to predict the likelihood of a double crossover from the individual probabilities of each single crossover
Interference
P (double crossover) = P (single crossover between b
and pr)
P (single crossover between
pr and vg)
X
= 0.061 X 0.123 = 0.0075
Based on a total of 1,005 offspring the expected number of double crossover offspring is
= 1,005 X 0.0075 = 7.5
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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
Interference
Interference (I) is expressed as I = 1 – C
where C is the coefficient of coincidence
C = Observed number of double crossovers
Expected number of double crossovers
3
7.5 C =
I = 1 – C = 1 – 0.4
= 0.6 or 60% This means that 60% of the expected number of
crossovers did not occur
= 0.40
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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
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
Other Mapping Techniques
Cytogenetic Mapping Have a mutant phenotype Perform karyotype Observe altered chromosomes
aneuploidies translocations deletions
Aberrant chromosome probablylocation of gene
normal
translocation t(14;17)
Patau Syndrome - Trisomy 13 Karyotype: 47, 13+
Partial Chromosomal Deletions
cri-du-chat - 46, 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
Mapping Strategies
Deletion Mapping Cross recessive heterozygote to
deletion heterozygote lines appearance of recessive phenotype
localizes recessive gene within deletion interval
Complementation
Determines if two phenotypes are caused by same mutant gene
Complementation Analysis
Complementation Analysis
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
Complementation Analysis
Mapping with Unknown Phase of LinkageWhen you have heterozygotes, but do not know what the
phenotypes of the parents were.
12 possible diploid arrangements for 3 linked locibm pr
bm pr
+ +
bm
+
bm
+
pr
+
brown midrib, virescent seedling, purple aleurone
Use Results of Cross to Determine Phase of Linkage
NCOs = parental chromosomes = phase
DCOs = fewestDo not directly detect middle locus
NCO Gives 3 Possibilities for Phase
12 possible diploid arrangements for 3 linked loci
bm pr
bm pr
+ +
bm
+
bm
+
pr
+
DCO Gives Loci Order
DCO Defines 1 of these 3 Possibilities for Phase
12 possible diploid arrangements for 3 linked loci
bm pr
bm pr
+ +
bm
+
bm
+
pr
+
Because a DCO was required to group all mutant loci on same homologue