1. Chapter 06 *Lecture Outline
*See separate FlexArt PowerPoint slides for all figures and tables pre-inserted into PowerPoint without notes.
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2. INTRODUCTION
Each species of organism must contain hundreds to thousands of genes
Yet most species have at most a few dozen chromosomes
Therefore, each chromosome is likely to carry many hundred or even thousands of different genes
The transmission genes close to one another on the same chromosome will violate Mendel’s law of independent assortment
6-2
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3. 6.1 LINKAGE AND CROSSING OVER
In eukaryotic species, each linear chromosome contains a long piece of DNA
A typical chromosome contains many hundred or even a few thousand different genes
The term synteny means two or more genes are located on the same chromosome and are physically linked
Genetic Linkage is the phenomenon that genes close together on a chromosome tend to be transmitted as a unit, which influences inheritance patterns
6-3
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4. Chromosomes are called linkage groups
They contain a group of genes that are linked together
The number of linkage groups is the number of types of chromosomes of the species
For example, in humans
22 autosomal linkage groups
An X chromosome linkage group
A Y chromosome linkage group
Genes that are far apart on the same chromosome may independently assort from each other
This is due to crossing over
A dihybrid cross studies linkage between two genes
A trihybrid cross studies linkage between three genes
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5. Crossing Over May Produce Recombinant Phenotypes
In diploid eukaryotic species, linkage can be altered during meiosis as a result of crossing over
Crossing over
Occurs during prophase I of meiosis
replicated sister chromatid homologues associate as bivalents
Non-sister chromatids of homologous chromosomes exchange DNA segments
Figure 6.1 illustrates the consequences of crossing over during meiosis
6-5
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6. (a) Without crossing over, linked alleles segregate together.
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The haploid cells contain the same combination of alleles as the original chromosomes B B b b A A a a
The arrangement of linked alleles has not been altered
Diploid cell after
chromosome replication
Meiosis B B A A b b a a
Possible haploid cells
(a) Without crossing over, linked
alleles segregate together.
Figure 6.1
6-6

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These haploid cells contain a combination of alleles NOT found in the original chromosomes B b A a
This new combination of alleles is a result of genetic recombination
These are termed parental or non-recombinant cells
Diploid cell after
chromosome replication
These are termed nonparental or recombinant cells
Meiosis B B A a b b a A
Possible haploid cells
(b) Crossing over can reassort
linked alleles.
Figure 6.1
6-7

8. Bateson and Punnett Discovered Two Traits That Did Not Assort Independently
In 1905, William Bateson and Reginald Punnett conducted a cross in sweet pea involving two different traits
Flower color and pollen shape
This is a dihybrid cross that is expected to yield a 9:3:3:1 phenotypic ratio in the F2 generation
However, Bateson and Punnett obtained surprising results
Refer to Figure 6.2
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9. Figure 6.2
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P generation x
Purple flowers,
long pollen (PPLL)
Red flowers,
round pollen (ppll )
A much greater proportion of the two types found in the parental generation
F1 offspring
Purple flowers,
long pollen (PpLl )
Self-fertilization
Observed
number
Expected
number
F2 offspring
Ratio
Ratio
Purple flowers, long pollen
296
15.6
240 9
Purple flowers, round pollen 19
1.0 80 3
Red flowers, long pollen 27
1.4 80 3
Red flowers, round pollen 85
4.5 27 1
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10. Bateson and Punnett Discovered Two Traits That Did Not Assort Independently
They suggested that the transmission of the two traits from the parents was somehow coupled
The two traits are not easily assorted in an independent manner
However, they did not realize that the coupling is due to the linkage of the two genes on the same chromosome
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11. Morgan Provided Evidence for the Linkage of Several X-linked Genes
The first direct evidence of linkage came from studies conducted by Thomas Hunt Morgan
Morgan investigated several traits that followed an X-linked pattern of inheritance in Drosophila
Figure 6.3 illustrates an experiment involving three traits
Body color
Eye color
Wing length
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12. 6-12 Figure 6.3 P generation x Xywm Xywm Xy+w+m+ Y F1 generation
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P generation x
Xywm Xywm
Xy+w+m+ Y
F1 generation
F1 generation
contains wild-type
females and
yellow-bodied,
white-eyed,
miniature-winged
males. x
Xy +w +m+ Xywm
Xywm Y
F2 generation
Females
Males
Total
Gray body, red eyes, long wings
439
319
758
Gray body, red eyes, miniature wings
208
193
401
Gray body, white eyes, long wings 1 1
Gray body, white eyes, miniature wings 5 11 16
Yellow body, red eyes, long wings 7 5 12
Yellow body, red eyes, miniature wings
Yellow body, white eyes, long wings
178
139
317
Yellow body, white eyes, miniature wings
365
335
700
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13. 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
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14. Morgan Provided Evidence for the Linkage of Several X-linked Genes
However, Morgan still had to interpret two key observations
1. Why did the F2 generation have a significant number of nonparental combinations?
2. Why was there a quantitative difference between the various nonparental combinations?
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15. White eyes, miniature wings 716 Red eyes, miniature wings 401
Let’s reorganize Morgan’s data by considering the 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
6-15
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16. Janssens had observed chiasmata microscopically
To explain these data, Morgan considered the previous studies of the cytologist F.A. Janssens
Janssens had observed chiasmata microscopically
And proposed that crossing over involves a physical exchange between homologous chromosomes
Morgan shrewdly realized that crossing over between homologous X chromosomes was consistent with his data
He assumed crossing over did not occur between the X and Y chromosome
The three genes were not found on the Y chromosome
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17. Morgan made three important hypotheses to explain his results
1. The genes for body color, eye color and wing length are all located on the X-chromosome
They tend to be inherited together
2. Due to crossing over, the homologous X chromosomes (in the female) can exchange pieces of the chromosomes
This created new combinations of alleles
3. The likelihood of crossing over depends on the distance between the two genes
Crossing over is more likely to occur between two genes that are far apart from each other
Figure 6.5 illustrates how crossing over provides an explanation for Morgan’s trihybrid cross
6-17
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18. Figure 6.5
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(a) No crossing over, parental offspring
(b) Crossover between eye color and wing length genes, fairly common
These parental phenotypes are the most common offspring
These recombinant offspring are not uncommon
because the genes are far apart
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19. 6-19 Figure 6.5 These recombinant offspring are fairly uncommon
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Figure 6.5
(c) Crossover between body color and eye color genes, uncommon
(d) Double crossover, very uncommon
These recombinant offspring are fairly uncommon
These double recombinant offspring are very unlikely,
1 out of 2,205
because the genes are
very close together
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20. Chi Square Analysis
This statistical method is frequently used to determine if the outcome of a dihybrid cross is consistent with linkage or independent assortment
Let’s consider the data concerning body color and eye color
An example of a chi square approach to determine linkage is shown next
6-20
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21. Step 1: Propose a hypothesis
The genes for eye color and body color are assorting independently
Even though the observed data appear inconsistent with this hypothesis, it allows us to calculate expected values
Since we do not know the probability of crossing over, we cannot calculate an expected value for a hypothesis based on linkage
Indeed, we actually anticipate that the chi square analysis will allow us to reject this hypothesis in favor of a linkage hypothesis
Step 2: Based on the hypothesis, calculate the expected values of each of the four phenotypes
An independent assortment hypothesis predicts that each phenotype has an equal probability of occurring
Refer to Punnett square on the next slide
6-21
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22. 6-22 F1 male gametes Xyw Y Xy+w+Xyw Xy+w+Y Xy+w+ Gray body, red eyes
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F1 male gametes
Xyw Y
Xy+w+Xyw
Xy+w+Y
Xy+w+
Gray body,
red eyes
Gray body,
red eyes
Xy+wXyw
Xy+wY
Xy+w
Gray body,
white eyes
Gray body,
white eyes
F1 female gametes
Xyw+Xyw
Xyw+Y
Xyw+
Yellow body,
red eyes
Yellow body,
red eyes
XywXyw
XywY
Xyw
Yellow body,
white eyes
Yellow body,
white eyes
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23. Data from the cross is shown in the table Total offspring equals 2,205
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
Data from the cross is shown in the table
Total offspring equals 2,205
Therefore, the expected number of each phenotype (combining males and females) is 1/4 X 2,205 = 551
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24. Step 3: Apply the chi square formula
(O1 – E1)2 E1
(O2 – E2)2 E2
(O3 – E3)2 E3
(O4 – E4)2 E4
c2 = + + +
(1159 – 551)2
551
(17 – 551)2
551
(12 – 551)2
551
(1017 – 551)2
551
c2 = + + +
c2 =
c2 = 2,109.8
6-24
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25. Step 4: Interpret the calculated chi square value
This is done with a chi square table as shown in Chapter 2
There are four experimental categories (n = 4)
Therefore, the degrees of freedom (df) is n -1 = 3
The calculated chi square value is enormous
Thus, the deviation between observed and expected values is very large
According to Table 2.1, such a large deviation is expected to occur by chance alone less than 1% of time
Therefore, we reject the hypothesis that the two genes assort independently
In other words, we accept the hypothesis that the genes are linked
6-25
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26. 6-26

27. Creighton and McClintock Experiment
To obtain direct evidence that genetic recombination is due to crossing over, Harriet Creighton and Barbara McClintock used the following strategy
1. They made crosses involving two linked genes
These crosses yielded parental and recombinant offspring
2. They observed the chromosomes in the parents and offspring microscopically
The parental chromosomes had some unusual structural features
See Figure 6.6
They wanted to see if there was a correlation between
The occurrence of recombinant offspring
and
Microscopically observable exchanges in segments of homologous chromosomes
6-27
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28. Creighton and McClintock focused on the pattern of inheritance for traits in corn
In previous cytological examinations of corn, they found a chromosome number 9 that had two abnormalities
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Normal
chromosome 9
Abnormal
chromosome 9
Due to a translocation
Knob
Translocated
piece from
chromosome 8
(a) Normal and abnormal chromosome 9
Figure 6.6
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29. They also knew of two relevant genes
Creighton and McClintock insightfully realized that this abnormal chromosome 9 could be used in crossing over experiments
They also knew of two relevant genes
A gene for kernel color located near the knobbed end
C = Colored
c = colorless
A gene for kernel endosperm texture located near the translocated end
Wx = Starchy endosperm
wx = waxy endosperm
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30. These chromosomes are distinguishable from the parental chromosomes
Creighton and McClintock reasoned that a crossover involving a normal and an abnormal chromosome 9 would yield
A chromosome that had either a knob or a translocation, but not both
These chromosomes are distinguishable from the parental chromosomes
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. c Wx
Parental
chromosomes C wx
Crossing over c wx
Nonparental
chromosomes C Wx
(b) Crossing over between normal and abnormal
chromosome 9
Figure 6.6
6-30

31. Testing the Hypothesis
Offspring with nonparental phenotypes are the product of a crossover
This crossover created nonparental chromosomes via an exchange of segments between homologous chromosomes
Testing the Hypothesis
Refer to Figure 6.7
6-31
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32. 6-32 Figure 6.7 Experimental level Conceptual level 1. Tassel C c c c
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Experimental level
Conceptual level 1.
Cross the two strains described. The
tassel is the pollen-bearing structure, and
the silk (equivalent to the stigma and
style) is connected to the ovary. After
fertilization, the ovary will develop into
an ear of corn.
Tassel C c c c x x
Silk wx Wx Wx wx
Parent A
Cc Wxwx
Parent B
cc Wxwx 2.
Observe the kernels from this cross. F1
ear of corn
Each kernel is a separate seed that has
inherited a set of chromosomes from
each parent. F1
kernels 3.
Microscopically examine chromosome 9
in the kernels.
Colored/waxy
Colorless/waxy
From parent B C c c c
A recombinant
chromosome wx wx wx wx
Microscope
From parent A
This illustrates only 2 possible
outcomes in the F1 kernels. The
recombinant chromosome on the right is
due to crossing over during meiosis in
parent A. As shown in The Data, there are
several possible outcomes.
Figure 6.7
6-32

33. The Data 6-33 Did a Crossover Occur During Gamete Formation
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Did a Crossover
Occur During
Gamete Formation
in Parent A?
Number of
Kernels
Analyzed
Phenotype of F
Kernel
Cytological Appearance of Chromosome 9 in F1 Offspring* 1
Colored/waxy 3
Knobbed/translocation
Normal No C wx c wx
Colorless/starchy 11
Knobless/normal
Normal No c Wx c or Wx c wx
Colorless/starchy 4
Knobless/translocation
Normal
Yes c wx c Wx
Colorless/waxy 2
Knobless/translocation
Normal
Yes c wx c wx
Normal
Colored/starchy 5
Knobbed/normal
Yes c or Wx C Wx
Total 25 c wx
*In this table, the chromosome on the left was inherited from parent A, and the blue chromosome on the right was inherited from parent B.
Data from Harriet B. Creighton and Barbara McClintock (1931) A Correlation of Cytological and Genetical Crossing-Over in Zea Mays. Proc. Natl. Acad. Sci.
USA 17, 492–497
6-33

34. Interpreting the Data
Creighton and McClintock were interested in whether crossing over occurred in parent A
It was the one heterozygous for both genes
The types of gametes produced by the parents are as follows
Parent A
Parent B
C wx (nonrecombinant)
c Wx (nonrecombinant)
C Wx (recombinant)
c wx (recombinant)
c Wx
c wx
By combining these gametes into a Punnett square, the following types of offspring can be produced
6-34
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35. So let’s start by considering the unambiguous phenotypes
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Parent B
c Wx
c wx
Ambiguous phenotypes that could be produced whether or not recombination occurred in parent A
Cc Wxwx
Cc wxwx
C wx
Nonrecombinant
Colored,
starchy
Colored,
waxy
cc WxWx
cc Wxwx
c Wx
Nonrecombinant
Colorless,
starchy
Colorless,
starchy
So let’s start by considering the unambiguous phenotypes
Cc WxWx
Cc Wxwx
C Wx
Recombinant
Colored,
starchy
Colored,
starchy
cc Wxwx
cc wxwx
c wx
Recombinant
Colorless,
starchy
Colorless,
waxy
6-35
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36. The colored, waxy phenotype (Cc wxwx) can occur only if Genetic Recombination did not occur between C and wx in parent A
If there was also no physical exchange, then parent A would pass the knobbed, translocated chromosome to its offspring
The colorless, waxy phenotype (cc wxwx) only occurs if Recombination did occur in parent A
The exchange would create a chromosome that had a translocation but was knobless
There was a perfect correlation between genetic recombination of alleles and the cytological presence of physical markers on the chromosomes
6-36
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37. As stated by Creighton and McClintock:
These observations were consistent with the idea that a crossover occurred between the C and wx genes involving a physical exchange of segments between homologous chromosomes
As stated by Creighton and McClintock:
“Pairing chromosomes, heteromorphic in two regions, have been shown to exchange parts at the same time they exchange genes assigned to these regions.”
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38. 6.2 GENETIC MAPPING IN PLANTS AND ANIMALS
Genetic mapping is also known as gene mapping or chromosome mapping
Its purpose is to determine the linear order of linked genes along the same chromosome
Figure 6.8 illustrates a simplified genetic linkage map of Drosophila melanogaster
6-38
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39. Copyright © The McGraw-Hill Companies, Inc
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Mutant
phenotype
Wild-type
phenotype
Mutant
phenotype
Wild-type
phenotype
Mutant
phenotype
Wild-type
phenotype
Mutant
phenotype
Wild-type
phenotype
Yellow
body, y
Gray
body
0.0
0.0
Aristaless, al
Long
aristae
0.0
Roughoid
eyes, ru
Smooth
eyes
1.5
White
eyes, w
Red
eyes
1.4
Bent
wings, bt
Straight
wings
13.0
Dumpy
wings, dp
Long
wings
26.0
Sepia
eyes, se
Red
eyes
3.0
Shaven
bristles, sv
Long
bristles
Vermilion
eyes, v
Red
eyes
33.0
36.1
Miniature
wings, m
Rudimentary
Long
wings 4
Long
wings
54.5
Black
body, b
Gray
body
wings, r
Bar
48.5
44.0
Scarlet
eyes, st
Red
eyes
Round
eyes
57.0
eyes, B
Carnation
54.5
Purple
eyes, pr
Red
eyes
62.5
Red
eyes
eyes, car
Bobbed
bristles, bb
58.5
Short
bristles, s
Long
bristles
66.0
Long
bristles
67.0
Vestigial
wings, vg
Long
wings
68.1
Little
fly, lf
Normal-
sized fly
Ebony
body, e
Gray
body
70.7
Each gene has its own unique locus at a particular site within a chromosome
1 (X)
75.5
Curved
wings, c
Straight
wings
Rough
eyes, ro
Smooth
eyes
91.1
100.7
Claret
eyes, ca
Red
eyes
104.5
Brown
eyes, bw
Red
eyes 2 3
Figure 6.8
6-39

40. Genetic maps are useful in many ways
1. They allow us to understand the overall complexity and genetic organization of a particular species
2. They can help molecular geneticists to clone genes
3. They improve our understanding of the evolutionary relationships among different species
4. They can be used to diagnose, and perhaps, someday to treat inherited human diseases
5. They can help in predicting the likelihood that a couple will produce children with certain inherited diseases
6. They provide helpful information for improving agriculturally important strains through selective breeding programs
6-40
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41. Number of recombinant offspring
Genetic maps allow us to estimate the relative distances between linked genes, based on the likelihood that a crossover will occur between them
Experimentally, the percentage of recombinant offspring is correlated with the distance between the two genes
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
6-41
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42. Figure 6.9 provides an example of a testcross
Genetic mapping experiments are typically accomplished by carrying out a testcross
A mating between an individual that is heterozygous for two or more genes and one that is homozygous recessive for the same genes
Figure 6.9 provides an example of a testcross
This cross concerns two linked genes affecting bristle length and body color in fruit flies
s = short bristles
s+ = normal bristles
e = ebony body color
e+ = gray body color
One parent displays both recessive traits
It is homozygous recessive for the two genes (ss ee)
The other parent is heterozygous for the two genes
The s and e alleles are linked on one chromosome
The s+ and e+ alleles are linked on the homologous chromosome
6-42
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43. Copyright © The McGraw-Hill Companies, Inc
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Chromosomes are the product of a crossover during meiosis in the heterozygous parent s+ s e+ e s s e e s+ s s s e+ e e e
Parent x
Parent s+ s s s s+ s s s e+ e e e e e e+ e
Recombinant offspring are fewer in number than nonrecombinant offspring s+ s
e + e s s e e s+ s e e s s e+ e
Long bristles
Gray body
Nonrecombinant
Short bristles
Ebony body
Nonrecombinant
Long bristles
Ebony body
Recombinant
Short bristles
Gray body
Recombinant
Figure 6.9
Total:
537
542 76 75
6-43

44. Number of recombinant offspring
The data at the bottom of Figure 6.9 can be used to estimate the distance between the two genes
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
6-44
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45. Alfred Sturtevant’s Experiment
The first genetic map was constructed in 1911 by Alfred Sturtevant
He was an undergraduate who spent time in the laboratory of Thomas Hunt Morgan
Sturtevant wrote:
“In conversation with Morgan … I suddenly realized that the variations in the strength 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.”
6-45
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46. Mutant allele (Phenotype) Wild-type allele (Phenotype)
Sturtevant considered the outcome of crosses involving six different mutant alleles
All of which are known to be recessive and X-linked
Mutant allele (Phenotype)
Wild-type allele (Phenotype)
y (yellow body color)
y+ (gray body color)
w (white eye color)
w+ (red eye color)
w-e (eosin eye color)
w-e+ (red eye color)
v (vermillion eye color)
v+ (red eye color)
m (miniature wings)
m+ (normal wings)
r (rudimentary wings)
r+ (normal wings)
Alleles of the same gene
Alleles of different genes that affect eye color
Alleles of different genes that affect wing length
Therefore, his genetic map contained only 5 genes
y, w, v, m and r
6-46
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47. Testing the Hypothesis
The distance between genes on a chromosome can be estimated from the proportion of recombinant offspring
This provides a way to map the order of genes along a chromosome
Testing the Hypothesis
Refer to Figure 6.10
6-47
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48. Copyright © The McGraw-Hill Companies, Inc
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Experimental level
Conceptual level
1. Cross a female that is heterozygous for
two different genes to a male that is
hemizygous recessive for the same two
genes. In this example, cross a female
that is Xy+w+Xyw to a male that is XywY.
This strategy was employed for many
dihybrid combinations of the six alleles
already described. y+ y y w+ w w x
Y chromosome
Offspring
2. Observe the outcome of the
crosses.
Parental types are more common y+ y y+
Gray bodies
Red eyes w+ w w+ y y y
Yellow bodies
White eyes w w w
Recombinant types are less common y+ y y+
Gray bodies
White eyes w w w y y y
Yellow bodies
Red eyes w+ w w+
3. Calculate the percentages of offspring
that are the result of crossing over
(number of recombinant/total).
See The Data.
Figure 6.10
6-48

49. The Data 6-49 Alleles Concerned Number Recombinant/ Total Number
Percent Recombinant Offspring
y and w/w-e
214/21,736
1.0
y and v
1,464/4,551
32.2
y and m
115/324
35.5
y and r
260/693
37.5
w/w-e and v
471/1,584
29.7
w/w-e and m
2,062/6,116
33.7
w/w-e and r
406/898
45.2
v and m
17/573
3.0
v and r
109/405
26.9
6-49
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50. Interpreting the Data
In some dihybrid crosses, the percentage of nonparental (recombinant) offspring was rather low
For example, there’s only 1% recombinant offspring in the crosses involving the y and w or w-e alleles
This suggests that these two genes are very close together
Other dihybrid crosses showed a higher percentage of nonparental offspring
For example, crosses between the v and r alleles produced 26.9% recombinant offspring
This suggests that these two genes are farther apart
6-50
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51. To construct his map, Sturtevant assumed that the map distances would be more accurate among genes that are closely linked
Therefore, his map is based on the following distances
y – w (1.0), w – v (29.7), v – m (3.0) and v – r (26.9)
Sturtevant also considered other features of the data to deduce the order of genes
For example,
Percentage of crossovers between w and m was 33.7
Percentage of crossovers between w and v was 29.7
Percentage of crossovers between v and m was 3.0
Therefore, v is between w and m, but closer to m
6-51
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52. Sturtevant began at the y gene and mapped the genes from left to right
Sturtevant collectively considered all these data and proposed the following genetic map
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29.7
23.9
1.0
3.0 y w v m r
0.0
1.0
30.7
33.7
57.6
Sturtevant began at the y gene and mapped the genes from left to right
6-52
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53. What is the basis for this inaccuracy?
A close look at Sturtevant’s data reveals two points that do not agree very well with his genetic map
The y and r dihybrid cross yielded 37.5% recombinants
But the map distance is 57.6
The w and r dihybrid cross yielded 45.2% recombinants
But the map distance is 56.6
What is the basis for this inaccuracy?
When the distance between two genes is large
The likelihood of multiple crossovers increases
Even numbers of crossover won’t be seen as recombination
So the observed number of recombinant offspring tends to underestimate the actual distance between the genes
As the percentage of recombinant offspring approaches 50%, map distances become progressively more inaccurate
6-53
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54. Copyright © The McGraw-Hill Companies, Inc
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 50
Percentage of recombinant
offspring in a testcross 25 10 50
100
150
Map units
Actual map distance along the chromosome
Figure 6.11
(computed from the analysis of many closely linked genes)
Multiple crossovers set a quantitative limit on measurable recombination frequencies as the physical distance increases
A testcross is expected to yield a maximum of only 50% recombinant offspring
6-54
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55. 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+ = gray body color
pr = purple eye color
pr+ = red eye color
vg = vestigial wings
vg+ = normal wings
6-55
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56. Step 1: Cross two true-breeding strains that differ with regard to three alleles.
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Parental flies
Male is homozygous wild-type for all three traits x
Female is mutant for all three traits
bb prpr vgvg
b+b+ pr+ pr+ vg+vg+ b pr vg b+
pr+
vg+ b pr vg b+
pr+
vg+
6-56

57. Step 2: Perform a testcross by mating F1 female heterozygotes to male flies that are homozygous recessive for all three alleles
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. x
b+b pr+pr vg+vg
F1 heterozygote
bb prpr vgvg
Homozygous recessive b+
pr+
vg+ b pr vg b pr vg b pr vg
During gametogenesis in the heterozygous female F1 flies, crossovers may produce new combinations of the 3 alleles
6-57
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58. Step 3: Collect data for the F2 generation
6-58

59. 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
6-59
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60. 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
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. b+
pr+
vg+ b+
pr+
vg+ b pr vg b
pr+ vg
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
6-60
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61. Step 4: Calculate the map distance between pairs of genes
To do this, one strategy is to regroup the data according to pairs of genes
From the parental generation, we know that the dominant alleles are linked, as are the recessive alleles
This allows us to group pairs of genes into parental and nonparental combinations
Parentals have a pair of dominant or a pair of recessive alleles
Nonparentals have one dominant and one recessive allele
The regrouped data will allow us to calculate the map distance between the two genes
6-61
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62. The map distance between body color and eye color is Map distance = 61
Parental offspring
Total
Nonparental Offspring
Gray body, red eyes
( )
472
Gray body, purple eyes
(30 + 2) 32
Black body, purple eyes
( )
Black body, red eyes
(28 + 1) 29
944 61
The map distance between body color and eye color is
Map distance = 61
X 100
= 6.1 map units
6-62

63. The map distance between body color and wing shape is Map distance =
Parental offspring
Total
Nonparental Offspring
Gray body, normal wings
( )
413
Gray body, vestigial wings
( ) 91
Black body, vestigial wings
( )
Black body, normal wings
( ) 88
826
179
The map distance between body color and wing shape is
Map distance =
179
X 100
= 17.8 map units
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64. The map distance between eye color and wing shape is Map distance =
Parental offspring
Total
Nonparental Offspring
Red eyes, normal wings
( )
439
Red eyes, vestigial wings
(61 + 1) 62
Purple eyes, vestigial wings
( )
442
Purple eyes, normal wings
(60 + 2)
881
124
The map distance between eye color and wing shape is
Map distance =
124
X 100
= 12.3 map units
6-64

65. 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
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12.3
6.1 b pr vg
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
6-65
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66. 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
6-66
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67. 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
6-67
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68. 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
6-68
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69. 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
6-69
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70. 6.3 GENETIC MAPPING IN HAPLOID EUKARYOTES
Much of our earliest understanding of genetic recombination came from the genetic analyses of fungi
Fungi may be unicellular or multicellular organisms
They are typically haploid (1n)
Haploid cells reproduce asexually
Two haploid cells can fuse to form a diploid zygote (2n) which goes through meiosis to produce haploid spores
The sac fungi (ascomycetes) have been particularly useful to geneticists because of their unique style of sexual reproduction
Refer to Figure 6.12
6-70
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71. 6-71 Figure 6.12 Meiosis produces four haploid cells, termed spores.
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Haploid cell (1n)
Haploid cell (1n)
Diploid zygote (2n)
Chromosome
replication
Meiosis produces four haploid cells, termed spores.
Meiosis
The group of four spores is known as a tetrad.
The spores are enclosed in a sac termed an ascus
Spore
Tetrad of haploid (1n)
spores contained within an ascus
Mitosis
(only certain species)
Some species form an octad of spores following a mitotic division
Spore
Figure 6.12
Octad of haploid (1n) spores contained
within an ascus
6-71

72. The cells of a tetrad or octad are contained within a sac
In other words, the products of a single meiotic division are contained within one sac
This is a key feature that dramatically differs from sexual reproduction in animals and plants
In animals, for example
Oogenesis only produces a single functional egg
Spermatogenesis produces sperm that are mixed with millions of other sperm
Using a microscope, researchers can dissect asci and study the traits of each haploid spore
The analysis of these asci can be used to map genes
6-72
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73. Types of Tetrads or Octads
The arrangement of spores within an ascus varies from species to species
Unordered tetrads or octads
Ascus provides enough space for the spores to randomly mix together
Ordered tetrads or octads
Ascus is very tight, thereby preventing spores from randomly moving around
Refer to Figure 6.13a
6-73
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74. 6-74 Tight ascus prevents mixing of spores
Ascus provides space for spores to randomly mix together
Saccharomyces
cerevisiae
Chlamydomonas
reinhardtii
Aspergillus
nidulans
Neurospora
crassa
Mold
Yeast
Unordered octad
Ordered octad
Unordered tetrads
Unicellular alga
(a) Different arrangements of spores
6-74
Figure 6.13
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75. Ordered Tetrad Analysis
Ordered tetrads or octads have the following key feature
The position and order of spores within the ascus is determined by the divisions of meiosis and mitosis
This idea is schematically shown in Figure 6.13b
The example depicts ordered octad formation in Neurospora crassa
Spores that carry the A allele show orange pigmentation
Spores that carry the a (albino) allele are white
6-75
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76. Pairs of daughter cells are located next to each other
All eight cells are arranged in a linear, ordered fashion
Pairs of daughter cells are located next to each other A A A A A A A A A A A a
Replication a
Meiosis I a
Meiosis II
Mitosis a a a a a a a
(b) Formation of an ordered octad in N. crassa a
Figure 6.13
6-76
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77. The genetic content of spores in ordered tetrads can be determined
This allows experimenters to map the distance between a single gene and the centromere
The logic of this mapping technique is based on the following features of meiosis
Centromeres of homologous chromosomes separate during meiosis I
Centromeres of sister chromatids separate during meiosis II
Refer to Figure 6.14
6-77
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78. 6-78 Figure 6.14 (a) No crossing over
Octad contains a linear arrangement of 4 haploid cells with the A allele which are adjacent to 4 with the a allele
This 4:4 arrangement of spores within the ascus is termed a first-division segregation (FDS) or an M1 pattern
Because the A and a alleles have segregated from each other after meiosis I A A A A A A 4 A A A A
Meiosis I
Meiosis II
Mitosis a a a a a a a 4 a
(a) First-division segregation (FDS)
No crossing over produces a 4:4 arrangement. a a
Figure 6.14 (a) No crossing over
6-78
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79. 6-79 Figure 6.14 (b) Single crossoverA A 2 A A A a a 2 A a a
Meiosis I
Meiosis II
Mitosis a A A A 2 A a a a a
These arrangements of spores are termed a second-division segregation (SDS) or M2 patterns 2
The A and a alleles do not segregate until meiosis II a A 2 A A A A a a A a a 4
Meiosis I
Meiosis II
Mitosis a a a a a a A A A
(b) Second-division segregation (SDS)
A single crossover can produce a 2:2:2:2 or 2:4:2 arrangement. 2 A
Figure 6.14 (b) Single crossover
6-79

80. The percentage of M2 asci can be used to calculate the map distance between the centromere and the gene of interest
The logic is that a gene is separated from its original centromere only after a crossover in the region between the gene and the centromere
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Centromere A
Centromere A A A A A a A A a a a a a a a
Result: The gene is
separated from its
original centromere.
Result: The gene is
not separated from its
original centromere.
Crossover begins between centromere
and gene of interest.
(b) Crossover does not begin between
centromere and gene of interest.
6-80
Figure 6.15

81. (1/2) (Number of SDS asci) X 100 Map distance = Total number of asci
Therefore the chances of getting a 2:2:2:2 or 2:4:2 pattern depend on the distance between the gene of interest and the centromere
To calculate this distance, the experimenter must count the number of SDS asci, as well as the total number of asci
In SDS asci, only half of the spores are actually the product of a crossover
Therefore
(1/2) (Number of SDS asci)
X 100
Map distance =
Total number of asci
6-81
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82. Unordered Tetrad Analysis
Unordered tetrads contain randomly arranged groups of spores
An experimenter can do a dihybrid cross and then determine the phenotypes of the spores
Such an analysis can determine if two genes are linked or assort independently
It can also be used to compute distance between two linked genes
6-82
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83. Unordered Tetrad Analysis
Consider a diploid yeast zygote with the genotype ura+ura-2 arg+arg-3
ura+ and arg+ = Normal alleles required for uracil and arginine biosynthesis, respectively
ura-2 and arg-3 = Defective alleles
Result in strains that require uracil and arginine in their growth medium
Figure 6.16 illustrates the assortment of the two genes in the unordered tetrad
6-83
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84. 6-84 Figure 6.16 PD Parental Ditype T Tetratype NPD Nonparental Ditype
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Haploid cell
ura +
arg x
ura -
2 arg-3
Haploid cell
PD Parental Ditype
T Tetratype
NPD Nonparental Ditype
ura + - 2
arg 3
Diploid zygote
Meiosis
Possible asci:
ura +
arg
ura-2 arg-3
ura +
arg
ura-2 arg-3
arg-3
ura-2 arg
ura +
arg-3
ura-2 arg
Parental ditype (PD)
Tetratype (T)
Nonparental ditype (NPD) 2
ura +
arg
: 2 -
2 arg 3 1
ura +
arg
: 1 -
2 arg 3 : 2
ura +
arg-3
: 2 -
2 arg
PD ascus: contains 100% parental cells
T ascus: contains 50% parental cells and 50% recombinant cells
NPD ascus: contains 100% recombinant cells
6-84

85. If the two genes assort independently
The number of asci with a parental ditype is expected to equal the number with a nonparental ditype
Thus, 50% recombinant spores are produced
If the two genes are linked
The type of crossover between them determines what type of ascus is produced
No crossovers yield the parental ditype
Single crossovers produce the tetratype
Double crossovers can yield any of the three types
The actual type produced depends on the combination of chromatids that are involved
Refer to Figure 6.17
6-85
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86. 6-86 Figure 6.17 (a) No crossing over (b) A single crossover
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Parental ditype
Tetratype
(a) No crossing over
(b) A single crossover 1 2 1 2
Nonparental ditype
Tetratype
Involving 4 sister chromatids
Involving 3 chromatids 1 2 1 2
Tetratype
Parental ditype
Involving 3 chromatids
Involving 2 chromatids
(c) Double crossovers
6-86

87. A more precise way to calculate map distance
As in conventional mapping, the map distance is calculated as the % of offspring that carry recombinant chromosomes
NPD + (1/2) (T)
Map distance =
X 100
Total number of asci
This calculation is fairly reliable over a short distance
However, over long distances it is not because it does not adequately account for double crossovers
A more precise way to calculate map distance
Single crossover tetrads +
(2) (Double crossover tetrads)
Map distance =
X 0.5 X 100
Total number of asci
Crossover tetrads also contain 50% nonrecombinant chromosomes
6-87
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88. So let’s take another look at Figure 6.17
For the equation to be useful, it needs to be related to the number of various types obtained by experimentation
So let’s take another look at Figure 6.17
The parental ditype (PD) and tetratype (T) are ambiguous
They can each be derived in two different ways
The nonparental ditype (NPD), however, is unambiguous
It can only be produced from a double crossover (DCO)
1/4 of all the double crossovers are nonparental ditypes
Therefore, DCO = 4 X NPD
But what about single crossovers (SCO)?
Notice that T asci can result from SCO or DCO
Since there are two kinds of T that are due to DCO
The actual number of T arising from DCO is 2NPD
So, T = SCO + 2NPD
Therefore, SCO = T – 2NPD
6-88
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89. Now we have accurate measures of both SCO and DCO
SCO = T – 2NPD and DCO = 4NPD
So, let’s substitute these values into our previous equation
Single crossover tetrads +
(2) (Double crossover tetrads)
Total number of asci
X 0.5 X 100
Map distance =
(T – 2NPD) + (2) (4NPD)
Map distance =
Total number of asci
X 0.5 X 100
T + 6NPD
Map distance =
Total number of asci
X 0.5 X 100
A more accurate measure of map distance because the equation considers both single- and double-crossovers
6-89
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90. 6.4 Mitotic Recombination
Mitosis does not involve the homologous pairing of chromosomes to form bivalents
Therefore crossing over in mitosis is expected to be much less likely than during meiosis
Nevertheless, crossing over does occur on rare occasions
In these cases, it may produce a pair of recombinant chromosomes that have a new combination of alleles
This is known as mitotic recombination
6-90
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91. Crossing Over Occasionally Occurs During Mitosis
If mitotic recombination occurs during an early stage of embryonic development
The daughter cells containing the recombinant chromosomes continue to divide
This may ultimately result in a patch of tissue with characteristics that are different from those of the rest of the organism
Curt Stern proposed that unusual patches on the bodies of certain Drosophila strains were due to mitotic recombination
6-91
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92. sn = short body bristles (singed) sn+ = normal body bristles
Stern was working with strains carrying X-linked genes affecting body color and bristle morphology
y = yellow body color
y+ = gray body color
sn = short body bristles (singed)
sn+ = normal body bristles
Females that are y+y sn+sn are expected to have gray body and normal bristles
However, when he microscopically observed these flies, he noticed places in which two adjacent regions were different from the rest of the body and from each other
This is called a twin spot
Stern proposed that twin spots are due to a single mitotic recombination within one cell during embryonic development
Refer to Figure 6.18
6-92
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93. Copyright © The McGraw-Hill Companies, Inc
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X chromosome
composition of
fertilized egg y+ sn y+
sn+
Normal mitotic divisions
to produce embryo
These cells are y+y sn+sn
Rare mitotic recombination
in one embryonic cell
Sister
chromatids y+ sn y+ sn y
sn+ y
sn+ y+ sn y
sn+ y+ sn y
sn+
So the twin spot is surrounded by cells that have gray color and normal bristles
Mitotic
crossover
Embryonic cell
Subsequent separation
of sister chromatids y+ sn y
sn+ y
sn+ y
sn+
Cytokinesis produces 2
adjacent cells
Patch with gray color and singed bristles y+ sn y+ sn y
sn+ y
sn+
Patch with yellow color and normal bristles
Continued normal mitotic
divisions to produce adult fly
with a twin spot
Figure 6.18
6-93