Many more genes in any organism than chromosomes and therefore there needs to be many gene loci per

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Many more genes in any organism than chromosomes
and therefore there needs to be many gene loci
per chromosome. These genes are located on
chromosomes like beads on a string.
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Genetic maps

• Genetic maps determine the order of genes based
  on meiotic recombination. The exact locality of
  a gene is not determined, but the relative
  distance between genes. These relative distances
  are expressed as centiMorgan (cM) and do not
  correspond directly to the true distance between
  the genes.
• Cytogenetic (physical) maps identify the precise
  location of a gene on a chromosome. The distance
  between genes are indicated as basepairs bp.
  (basepairs), kb. (1 000 bases kilobp) or mb. (1
  000 000 basepairs megabp).

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• Let us assume that the following homozygous lines
  are crossed
• AABB x aabb
• F1 AaBb
• Cross the F1 back to a homozygous recessive line

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• F2 expect observed
• AB 25 32
• Ab 25 18
• aB 25 15
• ab 25 35

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• Parental phenotypes (AB ab) are present in more
  than 50 of the offspring, therefore genes A and
  B are linked. The parental types are present in
  67 of the offspring, therefore crossing over
  between the two genes occurred in 33 of the
  offspring. This implies a distance of 33cM
  between the genes.

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3-point

• Use three genes
• Cross two homozygous parents
• Cross the F1 back to a homozygous recessive for
  all three genes
• The parental phenotypes should be complementary
• The numbers of F2s with each phenotype are
  counted

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• There is a positive correlation between the
  number of offspring and the reliability of the
  experiment
• At least 500 F2s should be used to calculate the
  distances but it is better to use at least 1 000
  F2s

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• Suppose you cross AABBcc with aabbCC
  (complimentary gene compositions)
• Cross the F1s back to the homozygous recessive
  line (aabbcc) and count the number of F2s in
  each phenotypic class. Suppose you observe the
  following numbers

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• Phenotype observed
• ABC 148
• ABc 1828
• AbC 227
• aBc 217
• abC 1779
• abc 139

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• Arrange the phenotypes in such a way that
  complimentary phenotypes are together
• The phenotype with the highest occurrence are
  placed first
• Phenotype observed
• ABc 1828 Parental types
• abC 1779
• ABC 148 Cross over 1
• abc 139
• AbC 227 Cross over 2
• aBc 217
• TOTAL 4338

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• Determine whether the genes are linked. This is
  done by calculating the percentage of parental
  types.
• Phenotype o /
• ABc 1828 3607 PT
• abC 1779 83.1
• ABC 148 287 CO 1
• abc 139 6.6
• AbC 227 444 CO 2
• aBc 217 10.2
• TOTAL 4338

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• Because the parental types occur in more than 50
  of the F2s (83.1), the genes are linked
• Determine the sequence of the genes (ABC, ACB or
  BAC)
• ABC ACB BAC (theoretical sequences)
• ABc AcB BAc (PT in different TS)
• abC aCb baC (Other PT)
• ABC ACB BAC (Co 1 in TS)
• abc acb bac (Complimentary sequence)
• AbC ACb bAC (CO 2 in TS)
• aBc acB Bac (Complimentary sequence)

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• A B c A c B B A c (Parental type)
• a b C a C b b a C (Compl.
  PT)
• A cross over between the first and second gene
  results in
• A b C A C b B a C
• Determine whether these theoretical combinations
  occur in cross over type 1 or 2
• A b C (2) A C b (2) B a C
• Determine the theoretical implications of a cross
  over between the second and third genes
• A B c A c B B A c (Parental type)
• a b C a C b b a C (Compl. PT)
• A cross over between the second and third gene
  results in
• A B C A c b B A C (Already excluded
  so ignore this sequence)
• Determine whether this theoretical combinations
  occur in cross over types 1 or 2
• A B C (1) A c b (does not occur in either 1
  or 2, therefore it is an incorrect sequence)

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• The correct sequence is consequently ABC (because
  both cross over types are theoretically
  possible). Rewrite the table with the correct
  sequence
• Phenotype observed /
• ABc 1828 3607 Parental types
• abC 1779 83.1
• ABC 148 287 Cross over 1
• abc 139 6.6
• AbC 227 444 Cross over 2
• aBc 217 10.2
• TOTAL 4338

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• Make a schematic representation of the DNA with
  the genes indicated on it
• A B C
• Calculate the distance between the genes
• A and B occur together in PT
• Cross over between A and B number of cross
  over that occurred between them (i.e. cases where
  Ab or aB occur together)
• 10.2 ? 10.2cM
• B and C ? 6.6cM.
• A 10.2cM B 6.6cM C

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• In this example some phenotypic classes are
  absent (aBC and Abc are absent). They represent
  double cross over events (i.e. a cross over
  between A and B, as well as a cross over between
  B and C). The frequency of double cross over is
  very low. Calculations in cases where double
  cross over occur, is done in a similar fashion,
  with the exception that only the double cross
  over is used to determine the sequence of the
  genus.

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• Suppose we obtain the following results from gene
  mapping
• A 17.2cM B 11.6cM C
• A cross over between A and B occur in 17.2 of
  the cases ? probability 17.2/100 0.172
• Probability of a double cross over the
  probability of a cross over between A and B, as
  well as between B and C. Therefore 0.172 x 0.116
  0.02.
• Sometimes synapses on one point of a chromosome
  influence synapses on another point. To
  determine the degree of chiasm interference we
  calculate
• C (observed frequency of double cross-overs) /
  (expected frequency of double cross-overs)
• I 1 C,
• where C equals the coefficient of similarity and
  I represents the chiasm interference

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• Use the following linkage data to construct a
  chromosome map of these loci.
• Loci Map units
• a-b 10 b-c 6
• a-c 16 b-d 13
• a-d 3 b-e 2
• a-e 8

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• A cross between individuals with an Lg N
  phenotype an lg n phenotype produced F1
  individuals with an Lg N phenotype. The F1 were
  crossed with individuals with lg n phenotypes,
  The F2 follow
• Progeny Number
• Lg N 180
• Lg n 50
• lg N 60
• lg n 190
• Total 480
• a) Are these genes linked?
• b) If they are linked, what were the genotypes
  of the original parental individuals?

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• Several female Drosophila heterozygous for four
  mutations were testcrossed. The mutations and
  their map positions were engrailed (en) map
  position 62.0, scabrous (sca) 66.7, droopy (dr)
  71.2, and curved (cu) 75.5.
• a) If the genotype of the original heterozygotes
  was en sca dr c / , how many different
  phenotypic classes would you expect to find in
  the progeny?
• b) If 25,000 progeny were examined, how many
  progeny with an engrailed, droopy phenotype would
  you expect to find, if there is no interference?

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• This map shows the distances between three loci
  (v, b, and lg) in corn. If individuals with a v
  b lg / genotype were test-crossed, what
  classes of phenotypes would you expect to find in
  the progeny, and how may of each would you expect
  if you examined 1000 progeny?
• v_________b________lg
• 18 28

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• A tomato plant heterozygous for alleles at three
  loci (Aa Bb Cc) was testcrossed to an aa bb cc
  plant. From the progeny listed here, which loci
  would you say are linked?
• Phenotype Number
• ABC 2250
• ABc 2225
• AbC 240
• Abc 255
• aBC 260
• aBc 250
• abC 2250
• abc 2275

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• A wild-type Drosophila with gray body and
  straight wings was crossed with a black-bodied,
  curled winged fly. The F1 were all gray bodied
  with straight wings. The F1 were back-crossed to
  black bodied, curled-winged flies. The progeny
  were as shown here.
• Cross Progeny Number
• F1 female x Black curled male
• Black curled 762
• Black straight 210
• Gray curled 234
• Gray straight 784
• F1 male x Black curled female
• Black curled 1013
• Black straight 0
• Gray curled 0
• Gray straight 987
• a) What is the map distance between these loci?
• b) What do you conclude about crossing over in
  male Drosophila?

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• From a three-point testcross mapping experiment,
  the following gamete genotype frequencies were
  obtained
• XYZ 365 xyz 367
• xYz 110 XyZ 105
• xYZ 3 Xyz 4
• XYz 25 xyZ 21
• TOTAL 1000
• From these data, what can be said of the genes?
• a) All are unlinked.
• b) XYZ is the gene order.
• c) YZX is the gene order.
• d) YXZ is the gene order.
• e) Two of the genes are linked one is
  independently assorting.

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• Stern and Bridges crossed a stock carrying the
  dominant eye mutation Star on the 2nd chromosome
  to a stock homozygous for the 2nd chromosome
  recessive mutations aristaless and dumpy. The F1
  Star females were then backcrossed to homozygous
  aristaless dumpy males and the following
  phenotypes observed
• Phenotype Number
• aristaless dumpy 918
• star 956
• aristaless star 7
• dumpy 5
• aristales 132
• star dumpy 100
• a) What are the recombination distances and the
  order of loci for these three genes?
• b) What classes of phenotypes are missing, and
  why?

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• In maize recessive mutant genes bm (brown
  midrib), v (virescent seedling) and pr (purple
  aleurone) are known. A plant heterozygous for
  all these genes are crossed with a homozygous
  recessive one for these genes and the following
  results are observed
• Phenotype number
• 4
• bm 182
• v 1200
• pr 2237
• pr v 79
• pr bm 1195
• v bm 2230
• pr v bm 3
• Determine whether these genes are linked. If
  they are linked, what is the order of the genes
  and the distances between them? Does chiasm
  interference occur? What are the genotypes of
  the parental types?

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• In the nematode Caenorhabditis elegans, d
  (dumpy), u (uncoordinated) and k (knobby) are all
  recessive genes located on the same chromosome.
  Females heterozygous normal for all three traits
  were mated with dumpy, uncoordinated, knobby
  males and the following progeny were observed.
• Phenotypes Number
• dumpy uncoordinated knobby 3
• uncoordinated knobby 392
• knobby 34
• uncoordinated 61
• dumpy uncoordinated 32
• dumpy knobby 65
• dumpy 410
• wild-type 3
• Total 1000

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• In the heterozygous normal females, what is the
  cis/trans arrangement of the following pairs of
  genes d u d k u k? Determine whether
  these genes are linked. If they are linked, what
  is the order of the genes and the distances
  between them?

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• In Drosophila a dominant mutation, dichaete (D),
  and two recessive mutations, pink (p) and ebony
  (e) are possibly linked. The F1s obtained from
  a cross of a dichaete line with a pink and ebony
  line are crossed to a homozygous recessive line
  and the following results were obtained
• Phenotype number
• ebony 8
• pink 187
• dichaete 8240
• pink, ebony 8781
• dichaete, ebony 214
• dichaete, pink 4
• dichaete, pink, ebony 28
• wild type 31
• Determine whether these genes are linked. If
  they are linked, what is the order of the genes
  and the distances between them? Does chiasm
  interference occur?

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• Yellow versus gray body colour in D. melanogaster
  is determined by the alleles y and , vermilion
  versus wild-type eyes by the alleles v and , and
  singed versus straight bristles by the alleles sn
  and . When females heterozygous for each of
  these X-linked genes were test-crossed with
  yellow, vermilion, singed males, the following
  classes and numbers of progeny were obtained.
• Phenotypes Number
• yellow, vermilion, singed 53
• yellow, vermilion 108
• yellow, singed 331
• yellow 5
• vermilion, singed 3
• vermilion 342
• singed 95
• wildtype 63

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• What is the order of the three genes? Construct
  a linkage map with the genes in their correct
  order, and indicate the map distance between the
  genes. How does the frequency of double
  crossovers observed in this experiment compare
  with the frequency expected if crossing-over
  occurs independently in the two chromosome
  regions? Determine the coefficient of
  coincidence an the interference.

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Linkage studies

• Physical or Biochemical markers Isozymes, RFLPs,
  RAPDs, DAFs, AFLPs, STSs, Micro satellites
  (VNTRs), QTLs

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AB/ab X ab/ab
Complete linkage GT PT ½ AB/ab AB ½ ab/ab ab
Partial linkage GT PT gt¼ AB/ab AB gt¼
ab/ab ab lt¼ aB/ab aB lt¼ Ab/ab Ab
No linkage GT PT ¼ AB/ab AB ¼ ab/ab ab ¼
aB/ab aB ¼ Ab/ab Ab
Partial linkage more progeny have parental
phenotypes and genotypes than have recombinant
types
No linkage equal number of progeny have parental
and recombinant phenotypes and genotypes
Complete linkage progeny only have parental
phenotypes and genotypes
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• Restriction Fragment Length Polymorphisms (RFLPs)
• Short pieces of random DNA obtained by cutting
  with restriction enzymes.
• Detection by Southern blot.
• Linkage of RFLP to physical characteristic or
  disease.
• Maps for the human and mouse genomes and for
  important agricultural crops such as potato and
  wheat.

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• Random Amplified Polymorphic DNA
• RAPDs
• PCR based technique.

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Polymerase Chain Reaction (PCR)
Amplification
Denaturing
Primer annealing
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• Random Amplified Polymorphic DNA
• RAPDs
• PCR based technique.
• Random primers anneal amplification of pieces
  of DNA of different lengths.
• Faster than RFLP and less DNA needed.
• Comparable to DAFs and AFLPs
• DNA Amplified Fingerprinting
• Amplified Fragment Length Polymorphism

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Random Priming
Genomic DNA
Primers
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• Sequence Tagged Sites
• STSs
• Detected by PCR
• Use two primers between which certain region is
  amplified

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STS
Primers
Genomic DNA
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• Microsatellites
• VNTR
• Variable Number of Tandem Repeats
• Repeat sequences made up of very short sequence
  motifs f.e. TG/CA
• Repeated between 50 and 100 000 times in human
  genome.
• Variable with regard to copy number and the copy
  number polymorphism is used