Linkage and genetic mapping in Eukaryotes

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Linkage and genetic mapping in Eukaryotes
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Key questions

• For genes on the same chromosome, can new
  combinations of alleles be detected?
• By what cellular mechanism does new combinations
  happen?
• Can the frequency of new combinations be related
  to the location of genes on chromosome?
• Can we test if genes are on the same chromosome?

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Outline

• Introduction
• Discovery of linked genes
• Recombination
• Linkage maps

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INTRODUCTION

• Each species of organism must contain hundreds to
  thousands of genes
• Yet there is only a small number of chromosomes
• Therefore, each chromosome is likely to carry
  many hundred or even thousands of different genes
• The transmission of such genes will violate
  Mendels law of independent assortment

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INTRODUCTION

• 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 linkage has two related meanings
• 1. Two or more genes can be located on the same
  chromosome
• 2. Genes that are close together tend to be
  transmitted as a unit linked genes

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DISCOVERY OF LINKED GENES

• 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 9331 phenotypic ratio in the F2
  generation
• However, Bateson and Punnett obtained surprising
  results

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DISCOVERY OF LINKED GENES

• 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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DISCOVERY OF 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
• One of his experiment involved three traits
• Body color
• Eye color
• Wing length

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• Morgan observed a much higher proportion of the
  combinations of traits found in the parental
  generation

• Morgans explanation
• All three genes are located on the X chromosome
• Therefore, they tend to be transmitted together
  as a unit they do not assort independently

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Chi Square Analysis

• This method is frequently used to determine if
  the outcome of a dihybrid cross is consistent
  with linkage or independent assortment
• Lets consider the data concerning body color and
  eye color

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• Step 1 Propose a hypothesis.
• The genes for eye color and body color are
  X-linked and, somehow, independently assorting
• Even though the observed data appear inconsistent
  with this hypothesis, the hypothesis allow us to
  calculate expected values
• 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

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• Total offspring equals 2,205
• Therefore, the expected number of each phenotype
  is 1/4 X 2,205 551

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• Step 3 Apply the chi square formula

c2



c2



c2 670.9 517.5 527.3 394.1
c2 2,109.8
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• Step 4 Interpret the calculated chi square value
• This is done with a chi square table
• 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
• Such a large deviation is expected to occur by
  chance alone less than 0.5 of time (p0.005,
  df3, c212.838)
• Therefore, we reject the hypothesis that the two
  genes assort independently
• In other words, we conclude that the genes are
  linked

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RECOMBINATION
Morgan provided evidence for the linkage of
several x-linked genes. However, he 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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RECOMBINATION

• Lets reorganize Morgans data by considering
  the pairs of genes separately

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RECOMBINATION

• 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
• Crossing over did not occur between the X and Y
  chromosome
• The three genes were not found on the Y
  chromosome Heteromorphic pair

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RECOMBINATION

• In diploid eukaryotic species, linkage can be
  altered during meiosis as a result of crossing
  over
• Crossing over
• Occurs during prophase I of meiosis at the
  bivalent stage
• Non-sister chromatids of homologous chromosomes
  exchange DNA segments
• The breakage of two DNA molecules at the same
  position and their rejoining in two reciprocal
  nonparental combinations.

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The haploid cells contain the same combination of
alleles as the original chromosomes
The arrangement of linked alleles has not been
altered
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These haploid cells contain a combination of
alleles NOT found in the original chromosomes
This new combination of alleles is a result of
genetic recombination
These are termed parental or non-recombinant cells
These are termed nonparental or recombinant cells
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RECOMBINATION

• The genetic analysis of organisms whose four
  products of meiosis remain together in groups
  demonstrated that crossing-over occurs at the
  four chromatid stages (bivalent stage)

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RECOMBINATION

• The genetic analysis of organisms whose four
  products of meiosis remain together in groups
  demonstrated that crossing-over occurs at the
  four chromatid stages (bivalent stage)

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RECOMBINATION

• For any pair of homologous chromosomes, 2, 3 or 4
  chromatids can take part in crossing-over events
  in a single meiocyte

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LINKAGE MAPS

• Terminology
• 1. A slash separates two homologs
• A/a (omitted if 2 identical alleles A A/A)
• 2. Alleles on the same homolog are separated by
  a comma
• A,B/a,b
• 3. Alleles are written in the same order on each
  homolog
• 4. Different chromosome locations are indicated
  by a semicolon
• A/aB/b
• 5. Alleles on the same homolog cis conformation,
  alleles on different homologs trans conformation
• A,B/a,b (A and B) and (a and b) are in cis (A
  and b) and (a and B) are in trans
• A,b/a,B (A and b) and (a and B) are in cis (A
  and B) and (a and b) are in trans
• Xw/Ycn,bwird1/ird2

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LINKAGE MAPS

• 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 chromosomes
• This created new combination 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

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These parental phenotypes are the most common
offspring
These recombinant offspring are not uncommon
because the genes are far apart
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These recombinant offspring are fairly uncommon
because the genes are very close together
These recombinant offspring are very unlikely 1
out of 2,205
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LINKAGE MAPS

• Linkage or 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

• The units of distance are called map units (mu)
• They are also referred to as centiMorgans (cM) or
  Sturts

• One map unit is equivalent to 1 recombination
  frequency

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LINKAGE MAPS

• 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
• Testcross
• This cross concerns two linked genes affecting
  bristle length and body color in fruit flies

• e ebony body color
• e gray body color

• s short bristles
• s normal bristles

• 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

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Chromosomes are the product of a crossover during
meiosis in the heterozygous parent
Recombinant offspring are fewer in number than
nonrecombinant offspring
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LINKAGE MAPS

• The data can be used to estimate the distance
  between the two genes
• Map distance

76 75
X 100

542 537 76 75
12.3 map units

• Therefore, the s and e genes are 12.3 map units
  apart from each other along the same chromosome

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Alfred Sturtevants 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 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.

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• Sturtevant considered the outcome of crosses
  involving six different mutant alleles
• All of which are known to be recessive and
  X-linked

• Therefore, his genetic map contained only 5 genes
• y, w, v, m and r

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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

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The Data
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Interpreting the Data

• In some dihybrid crosses, the percentage of
  nonparental (recombinant) offspring was rather
  low
• For example, theres 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 m alleles
  produced 26.9 recombinant offspring
• This suggests that these two genes are farther
  apart

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• 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 r (3.0) and v m
  (26.9)
• Sturtevant also considered other features of the
  data to deduce the order of genes
• For example,
• Percentage of crossovers between w and r was 33.7
• Percentage of crossovers between w and v was 29.7
• Percentage of crossovers between v and r was 3.0
• Therefore, the gene order is w v r
• Where v is closer to r than it is to w

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• Sturtevant collectively considered all these data
  and proposed the following genetic map

• Sturtevant began at the y gene and mapped the
  genes from left to right

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• A close look at Sturtevants data reveals two
  points that do not agree very well with his
  genetic map
• The y and m dihybrid cross yielded 37.5
  recombinants
• But the map distance is 57.6
• The w and m dihybrid cross yielded 45.2
  recombinants
• But the map distance is 56.6
• So whats up?

• When the distance between two genes is large
• The likelihood of multiple crossovers increases
• This causes the observed number of recombinant
  offspring to underestimate the distance between
  the two genes

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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

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• Step 1 Cross two true-breeding strains that
  differ with regard to three alleles.

Male is homozygous wildtype for all three traits
Female is mutant for all three traits

• The goal in this step is to obtain aF1
  individuals that are heterozygous for all three
  genes

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• Step 2 Perform a testcross by mating F1 female
  heterozygotes to male flies that are homozygous
  recessive for all three alleles

• During gamete formation 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

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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
• Most of the time (therefore it is better to AVOID
  these rules), 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 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

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• The map distance between body color and eye color
  is
• Map distance

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X 100
6.1 map units
944 61
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• The map distance between body color and wing
  shape is
• Map distance

17.8 map units
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• The map distance between eye color and wing shape
  is
• Map distance

12.3 map units
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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

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• To calculate map distance, we have gone through a
  method that involved the separation of data into
  pairs of genes (see step 4)
• An alternative method does not require this
  manipulation
• Rather, the trihybrid data is used directly
• This method is described next

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b,pr,vg b,pr,vg
Single crossover between b and pr
0.058
b,pr,vg b,pr,vg
Single crossover between pr and vg
0.120
b,pr,vg b,pr,vg
Double crossover, between b and pr, and between
pr and vg
0.003
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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 20.003) X 100
• 18.4 mu

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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
• Note that in the first method (grouping in
  pairs), the distance between b and vg was found
  to be 17.8 mu.
• This slightly lower value was a small
  underestimate because the first method does not
  consider the double crossovers in the calculation

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Interference

• 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
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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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

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• Interference (I) is expressed as
• I 1 C
• where C is the coefficient of coincidence

• C

• 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

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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

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• In mice, the genes associated with two nervous
  disorders, waltzer (w) and jittery (j) are
  located on the same chromosome 10cM apart. An
  unlimited number of wild-type heterozygous mice
  wj/w j are being maintained by a commercial
  firm (w is recessive resulting in the waltzer
  phenotype, j is recessive resulting in the
  jittery phenotype, w and j are dominant
  resulting in wild-type phenotype). An order
  arrives for two dozen male waltzer mice
  heterozygous for j. The average litter size per
  female is 6 offspring. Including a 10 safety
  factor to ensure the recovery of the needed
  number of offspring, how many wj/w j females
  will you have to mate with wj/w j males to fill
  this order?

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• In mice, the genes associated with two nervous
  disorders, waltzer (w) and jittery (j) are
  located on the same chromosome 10cM apart. An
  unlimited number of wild-type heterozygous mice
  wj/w j are being maintained by a commercial
  firm (w is recessive resulting in the waltzer
  phenotype, j is recessive resulting in the
  jittery phenotype, w and j are dominant
  resulting in wild-type phenotype). An order
  arrives for two dozen male waltzer mice
  heterozygous for j. The average litter size per
  female is 6 offspring. Including a 10 safety
  factor to ensure the recovery of the needed
  number of offspring, how many wj/w j females
  will you have to mate with wj/w j males to fill
  this order?

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• In mice, the genes associated with two nervous
  disorders, waltzer (w) and jittery (j) are
  located on the same chromosome 10cM apart. An
  unlimited number of wild-type heterozygous mice
  wj/w j are being maintained by a commercial
  firm (w is recessive resulting in the waltzer
  phenotype, j is recessive resulting in the
  jittery phenotype, w and j are dominant
  resulting in wild-type phenotype). An order
  arrives for two dozen male waltzer mice
  heterozygous for j. The average litter size per
  female is 6 offspring. Including a 10 safety
  factor to ensure the recovery of the needed
  number of offspring, how many wj/w j females
  will you have to mate with wj/w j males to fill
  this order?

Parental
Parental
Recombinant
Recombinant
10cM parental 90 recombinant 10
Parental
Recombinant
Recombinant
Parental
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• In mice, the genes associated with two nervous
  disorders, waltzer (w) and jittery (j) are
  located on the same chromosome 10cM apart. An
  unlimited number of wild-type heterozygous mice
  wj/w j are being maintained by a commercial
  firm (w is recessive resulting in the waltzer
  phenotype, j is recessive resulting in the
  jittery phenotype, w and j are dominant
  resulting in wild-type phenotype). An order
  arrives for two dozen male waltzer mice
  heterozygous for j. The average litter size per
  female is 6 offspring. Including a 10 safety
  factor to ensure the recovery of the needed
  number of offspring, how many wj/w j females
  will you have to mate with wj/w j males to fill
  this order?

P0.02250.02250.045 4.5 of offsprings Males
2.25 If 100 offsprings give 2.25 males Then x
offsprings give 24 males x (100x24)/2.251066.7
1 female gives 6 offsprings x females gives
1066.7 offsprings x (1066.7x1)/6177.8 10 of
177.8 is 17.7 Answer 177.817.7196
Parental
Parental
Recombinant
Recombinant
Parental
Recombinant
Recombinant
Parental
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• Let consider three different genes. The recessive
  allele b gives flies a black body while the
  dominant b allele gives brown. The recessive
  allele wx gives waxy wings while the dominant wx
  allele gives nonwaxy wings. The recessive allele
  cn gives cinnabar eyes while the dominant cn
  allele gives red eyes. A female heterozygous for
  the three genes is testcrossed. 1000 progeny are
  obtained as follows
• Brown,nonwaxy,red 5 Black,waxy,cinnabar 6
• Brown,waxy,cinnabar 69 Black,nonwaxy,red 67
• Brown,nonwaxy,cinnabar 382 Black,waxy,red 379
• Brown,waxy,red 48 Black,nonwaxy,cinnabar 44
• 1. What is the genotype of the female?
• 2. Draw the genetic map for these genes
• 3. If appropriate, calculate interference

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• Let consider three different genes. The recessive
  allele b gives flies a black body while the
  dominant b allele gives brown. The recessive
  allele wx gives waxy wings while the dominant wx
  allele gives nonwaxy wings. The recessive allele
  cn gives cinnabar eyes while the dominant cn
  allele gives red eyes. A female heterozygous for
  the three genes is testcrossed. 1000 progeny are
  obtained as follows
• Brown,nonwaxy,red 5 Black,waxy,cinnabar 6
• Brown,waxy,cinnabar 69 Black,nonwaxy,red 67
• Brown,nonwaxy,cinnabar 382 Black,waxy,red 379
• Brown,waxy,red 48 Black,nonwaxy,cinnabar 44
• 1. What is the genotype of the female?
• Parental
• Brown,nonwaxy,cinnabar 382 Black,waxy,red 379
• Genotype b,wx,cn/b,wx,cn

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• Let consider three different genes. The recessive
  allele b gives flies a black body while the
  dominant b allele gives brown. The recessive
  allele wx gives waxy wings while the dominant wx
  allele gives nonwaxy wings. The recessive allele
  cn gives cinnabar eyes while the dominant cn
  allele gives red eyes. A female heterozygous for
  the three genes is testcrossed. 1000 progeny are
  obtained as follows
• Brown,nonwaxy,red 5 Black,waxy,cinnabar 6
• Brown,waxy,cinnabar 69 Black,nonwaxy,red 67
• Brown,nonwaxy,cinnabar 382 Black,waxy,red 379
• Brown,waxy,red 48 Black,nonwaxy,cinnabar 44
• 2. Draw the genetic map for these genes
• Cross b,wx,cn/b,wx,cn x b,wx,cn/b,wx,cn
• Distance b-wx
• Cross b,wx/b,wx x b,wx/b,wx
• parental recombinant
• Brown, nonwaxy 3825 Brown,waxy 6948
• Black,waxy 3796 Black,nonwaxy 6744
• d(6948)(6744)x100/100022.8mu
• Distance b-cn
• Cross b,cn/b,cn x b,wx/b,wx
• parental recombinant
• Brown, cinnabar 38269 Brown,red 548
• Black,red 37967 Black,cinnabar 644
• d5350/1010.3mu

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• Let consider three different genes. The recessive
  allele b gives flies a black body while the
  dominant b allele gives brown. The recessive
  allele wx gives waxy wings while the dominant wx
  allele gives nonwaxy wings. The recessive allele
  cn gives cinnabar eyes while the dominant cn
  allele gives red eyes. A female heterozygous for
  the three genes is testcrossed. 1000 progeny are
  obtained as follows
• Brown,nonwaxy,red 5 Black,waxy,cinnabar 6
• Brown,waxy,cinnabar 69 Black,nonwaxy,red 67
• Brown,nonwaxy,cinnabar 382 Black,waxy,red 379
• Brown,waxy,red 48 Black,nonwaxy,cinnabar 44
• 2. Draw the genetic map for these genes
• Cross b,wx,cn/b,wx,cn x b,wx,cn/b,wx,cn
• Distance b-wx
• Cross b,wx/b,wx x b,wx/b,wx
• parental recombinant
• Brown, nonwaxy 3825 Brown,waxy 6948
• Black,waxy 3796 Black,nonwaxy 6744
• d(6948)(6744)x100/100022.8mu
• Distance b-cn
• Cross b,cn/b,cn x b,wx/b,wx
• parental recombinant
• Brown, cinnabar 38269 Brown,red 548
• Black,red 37967 Black,cinnabar 644
• d5350/1010.3mu

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• Let consider three different genes. The recessive
  allele b gives flies a black body while the
  dominant b allele gives brown. The recessive
  allele wx gives waxy wings while the dominant wx
  allele gives nonwaxy wings. The recessive allele
  cn gives cinnabar eyes while the dominant cn
  allele gives red eyes. A female heterozygous for
  the three genes is testcrossed. 1000 progeny are
  obtained as follows
• Brown,nonwaxy,red 5 Black,waxy,cinnabar 6
• Brown,waxy,cinnabar 69 Black,nonwaxy,red 67
• Brown,nonwaxy,cinnabar 382 Black,waxy,red 379
• Brown,waxy,red 48 Black,nonwaxy,cinnabar 44
• 3. If appropriate, calculate interference
• crossover b-cn
• b cn wx
• b cn wx
• Brown, red, waxy 48
• Black, cinnabar, nonwaxy 44
• PCO b-cn(4844)/10000.092
• crossover cn-wx
• b cn wx
• b cn wx
• Brown, cinnabar, waxy 69
• Black, red, nonwaxy 67
• PCO cn-wx136/10000.136
• Double-crossover