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Recombination Frequencies

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					                  Assignments
• Read from Chapter 3, 3.6 (pp. 100-106),

• Master Problems…3.12, 3.15, 3.20,

• Chapter 4, Problems 1, 2,

• Questions 4.1 - 4.4, 4.6, 4.7, 4.9, 4.11 -4.14, 4.19 -
  4.20 a,b,c,d.

• Exam Week from Friday…
   – One hour (you can use the entire 80 minutes, but no
     more). One 8” x 11”, one sided crib sheet.
    Sex Determination Systems

• Different mechanisms of sex selection exist:
                         »
   • XX / XO (O = null),

   • ZW / ZZ (female ZW, Male ZZ),

   • haplo / diplo (males are haploid),

   • XX / XY (most mammals).
          Sex Chromosomes
                 most mammals…



... ‘X’ and ‘Y’ chromosomes that determine
   the sex of an individual in many organisms,
                   Females: XX

                   Males: XY
                      a                            A                a

                          hemizygous:
                          condition where gene
  Differential            is present in only one
  Region                  dose (one allele).




                               Differential Region

Paring Region                  Paring Region


                 XY: male                              XX: female
                 X Linkage
…the pattern of inheritance resulting from genes
 located on the X chromosome.



X-Linked Genes…

…refers specifically to genes on the X-
 chromosome, with no homologs on the Y
 chromosome.
                               Blue is dominant.


P
                   x



           Blue        Pink
          Female       Male




Gametes
                          or
Gametes
                             or




 F1




          Blue Female   Blue Male
F1                      x




          Blue Female       Blue Male




Gametes
             or                or
     Gametes
                     or                       or




F2




      Blue Female   Blue Male   Blue Female        Pink Male
F2




     Blue Female    Blue Male   Blue Female   Pink Male



                   3 : 1 Blue to Pink

                   1 : 1 Female to Male
P
                   x



           Pink        Blue
          Female       Male




Gametes
                          or
Gametes
                             or




 F1




          Blue Female   Pink Male
     Gametes
                     or                       or




F2




      Pink Female   Pink Male   Blue Female        Blue Male
F2




     Pink Female    Pink Male   Blue Female   Blue Male


        1               1            1           1

                   1 : 1 Female to Male
                     1 : 1 Pink to Blue
        Sex Linkage to Ponder
• Female is homozygous recessive X-linked gene,
  – what percentage of male offspring will express?

  – what percentage of female offspring will express if,
     • mate is hemizygous for the recessive allele?
     • mate is hemizygous for the dominant allele?


• Repeat at home with female heterozygous X-
  linked gene!
     Sex-Linked vs. Autosomal
• autosomal chromosome: non-sex linked
  chromosome,

• autosomal gene: a gene on an autosomal
  chromosome,

• autosomes segregate identically in reciprocal
  crosses.
      X-Linked Recessive Traits
                             Characteristics


• Many more males than females show the phenotype,

   – female must have both parents carrying the allele,
   – male only needs a mother with the allele,


• Very few (or none) of the offspring of affected males show
  the disorder,

   – all of his daughters are carriers,

       • roughly half of the sons born to these daughters are carriers.
            X-Linked Dominant
• Affected males married to unaffected females pass the
  phenotype to their daughters, but not to their sons,

• Heterozygous females married to unaffected males pass the
  phenotype to half their sons and daughters,

• Homozygous dominant females pass the phenotype on to
  all their sons and daughters.
        Autosomal Dominant


• Phenotypes appear in every generation,

• Affected males and females pass the
  phenotype to equal proportions of their sons
  and daughters.
     Pedigree for Very Rare Trait
                   ? = kid with trait




                                              1/2


                                        1/2




Recessive?   ---> Yes!                         ?
                                   1/2 x 1/2 x 1/2 = 1/8 ?

                                                    x 1/2 = 1/16
X-Linked?
Autosomal?   ---> Yes!
                                          (p)boy
             X-Linked Dominant
                     examples (OMIM)

• HYPOPHOSPHATEMIA: “Vitamin-D resistant Rickett’s”,

• LISSENCEPHALY: “smooth brain”,

• FRAGILE SITE MENTAL RETARDATION: mild retardation,

• RETT Syndrome: neurological disorder,

• More on OMIM…
Genetics:
…in the News
              Linkage


• Genes linked on the same chromosome may
  segregate together.
        Independent Assortment

A   a       B       b
                                2n = 4




A               A       a   B                b
    B                                    a
                    b
                                              2n = 1
Meiosis               A   a
No Cross Over
                      B   b           Parent Cell




    A             A              a              a
    B             B              b              b



        Daughter Cells Have Parental Chromosomes
                                                2n = 1
Meiosis                  A   a
With Cross Over
                         B   b          Parent Cell




    A              A               a              a
    B              b               B              b



        Daughter Cells Have Recombinant Chromosomes
       Dihybrid Cross

                             yellow/round   phenotype
P           green/wrinkled
                                            genotype
     GGWW        x           ggww
                                            gametes
      GW                      gw

F1                                          genotype
            GgWw
    Gamate Formation in F1 Dihybrids
         P: GGWW x ggww, Independent Assortment



F1 Genotype:         GgWw


   G           g             W       w            alleles




   GW          Gw           gW      gw            gametes

   .25         .25          .25     .25           probability
          How do you test for
         assortment of alleles?

                        F1: GgWw


           GW          Gw         gW            gw
           .25         .25        .25          .25



Test Cross: phenotypes of the offspring indicate the genotype of
       the gametes produced by the parent in question.
               Test Cross
               GgWw x ggww


GW (.25)   x   gw (1)        GgWw (.25)


Gw (.25)   x   gw (1)        G gww (.25)


gW (.25)   x   gw (1)        ggWw (.25)


gw (.25)   x   gw (1)        ggww (.25)
                 Test Cross
                 GgWw x ggww


GW (.25)    x     gw (1)          GgWw (.25)   P

Gw (.25)    x     gw (1)          Ggww (.25)   R

gW (.25)    x     gw (1)          ggWw (.25)   R

gw (.25)    x     gw (1)          ggww (.25)   P

F1 parental types GgWw and gwgw

recombinant types Ggww and ggWw
    Recombination Frequency

…or Linkage Ratio: the percentage of
 recombinant types,

  – if 50%, then the genes are not linked,

  – if less than 50%, then linkage is observed.
                 Linkage

• Genes closely located on the same
  chromosome do not recombine,
  – unless crossing over occurs,


• The recombination frequency gives an
  estimate of the distance between the genes.
    Recombination Frequencies

• Genes that are adjacent have a recombination
  frequency near 0%,
• Genes that are very far apart on a chromosome
  have a recombination frequency of 50%,
• The relative distance between linked genes
  influences the amount of recombination observed.
                                                      homologs
 A          B



  a         b
In this example, there is a 2/10 chance of recombination.



  A                   C



  a                   c
In this example, there is a 4/10 chance of recombination.
                      Linkage Ratio
                       P GGWW x ggww

                   Testcross F1: GgWw x ggww

             GW           Gw           gW              gw

determine      ?            ?            ?              ?


            # recombinant x 100 = Linkage Ratio
            # total progeny
            Units: % = mu (map units) - or - % = cm (centimorgan)
                                             Study Figs 4.2, 4.3, and 4.5

Fly Crosses (simple 3-point mapping)
                (white eyes, minature, yellow body)


 • In a white eyes x miniature cross, 900 of the 2,441
   progeny were recombinant, yielding a map
   distance of 36.9 mu,

 • In a separate white eyes x yellow body cross, 11 of
   2,205 progeny were recombinant, yielding a map
   distance of 0.5 mu,

 • When a miniature x yellow body cross was
   performed, 650 of 1706 flies were recombinant,
   yielding a map distance of 38 mu.
              Simple Mapping
• white eyes x miniature = 36.9 mu,

• white eyes x yellow body = 0.5 mu,

• miniature x yellow body = 38 mu,
        0.5 mu

                        36.9 mu


        y w                            m
                        38 mu
  Do We have to Learn More Mapping
            Techniques?

• Yes,
  – three point mapping,

• Why,
  –   Certainty of Gene Order,
  –   Double crossovers,
  –   To answer Cyril Napp’s questions,
  –   and, for example: over 4000 known human diseases
      have a genetic component,
       • knowing the protein produced at specific loci facilitates the
         treatment and testing.
   cis
“coupling”
   trans
“repulsion”
                             Classical Mapping
                                                                         Cross an organism with
                                                                         a trait of interest to
                        target                                           homozygous mutants of
                                                                         known mapped genes.

                                                                         Then, determine if
                                                                         segregation is random in
                                                                         the F2 generation,
                                                                            • if not, then your
                                                                            gene is linked
                                                                            (close) to the known
What recombination frequency do you expect between the target and HY2?
What recombination frequency do you expect between the target and TT2?
                                                                            mapped gene.
                  Gene Order
• It is often difficult to assign the order of genes
  based on two-point crosses due to uncertainty
  derived from sampling error.

                   A x B = 37.8 mu,
                   A x C = 0.5 mu,
                   B x C = 37.6 mu,
            Double Crossovers
• More than one crossover event can occur in a
  single tetrad between non-sister chromatids,
   – if recombination occurs between genes A and B 30% of
     the time (p = 0.3), then the probability of the event
     occurring twice is 0.3 x 0.3 = 0.09, or nearly one map
     unit.

• If there is a double cross over, does recombination
  occur?
   – how does it affect our estimation of distance between
     genes?
                             Classical Mapping
                                              model organisms



                                                                         Cross an organism with
                                                                         a trait of interest to
                        target                                           homozygous mutants of
                                                                         known mapped genes.

                                                                         Then, determine if
                                                                         segregation is random in
                                                                         the F2 generation,
                                                                            • if not, then your
                                                                            gene is linked
                                                                            (close) to the known
What recombination frequency do you expect between the target and HY2?
What recombination frequency do you expect between the target and TT2?
                                                                            mapped gene.
Classical mapping in
humans requires
pedigrees…
  Three Point Testcross

    Triple Heterozygous
          (AaBbCc )


             x

Triple Homozygous Recessive
          (aabbcc)
 Three Point Mapping Requirements

• The genotype of the organism producing the gametes must
  be heterozygous at all three loci,

• You have to be able to deduce the genotype of the gamete
  by looking at the phenotype of the offspring,

• You must look at enough offspring so that all crossover
  classes are represented.
w   g   d
            w     g        d

Representing linked genes...

                 + + +
 P                             = WwGgDd
                 w g d

                    x

                 w g d
Testcross                      = wwggdd
                 w g d
Phenotypic Classes
     tri-hybrid cross?
           #
W-G-D-    179
                Parentals
wwggdd    173

W-G-dd    46    Recombinants
                1 crossover,
wwggD-    52    Region I

wwG-D-    22    Recombinants
                1 crossover,
W-gg-dd   22    Region II

W-gg-D     2
                Recombinants,
                double crossover
wwG-dd     4
            #
W-G-D-     179                                         I
                 Parentals
wwggdd     173                      W G               D
W-G-dd     46    Recombinants       w g           d
                 1 crossover,
wwggD-     52    Region I
                                    Region I:
wwG-D-     22    Recombinants
                 1 crossover,
W-gg-dd    22    Region II           46 + 52 + 2 + 4
                                                           x 100
W-gg-D       2                            500
                 Recombinants,
                 double crossover
wwG-dd      4
                                          = 20.8 mu
 Total =   500
            #
                                        II         20.8 mu
W-G-D-     179
                 Parentals
wwggdd     173                      W G               D
W-G-dd     46    Recombinants       w g            d
                 1 crossover,
wwggD-     52    Region I
                                    Region II:
wwG-D-     22    Recombinants
                 1 crossover,
W-gg-dd    22    Region II           22 + 22 + 2 + 4
                                                          x 100
W-gg-D       2                               500
                 Recombinants,
                 double crossover
wwG-dd      4
                                          = 10.0 mu
 Total =   500
 10.0 mu         20.8 mu

W G              D
                                        0.1 x 0.208 = 0.0208
w g          d

                                                          NO GOOD!

W-gg-D       2
                   Recombinants,
                   double crossover      6/500 = 0.012
wwG-dd      4

 Total =   500

                           Coefficient of Coincidence =    Observed
                                                           Expected
                           Interference = 1 - Coefficient of Coincidence
                Interference
…the effect a crossing over event has on a
 second crossing over event in an adjacent
 region of the chromatid,

  – (positive) interference: decreases the
    probability of a second crossing over,
     • most common in eukaryotes,

  – negative interference: increases the
    probability of a second crossing over.
Gene Order in Three Point Crosses

• Find - either - double cross-over
  phenotype…based on the recombination
  frequencies,

• Two parental alleles, and one cross over allele will
  be present,

• The cross over allele fits in the middle...
           #
A-B-C-    2001
aabbcc    1786

A-B-cc    46     Which one is the “odd”
aabbC-    52     one?

aaB-cc    990
                         II            I
A-bb-C-   887        A C           B
A-bb cc   600
                     a c          b
aaB-C-    589
           #     Region I
A-B-C-    2001
aabbcc    1786    990 + 887 + 46 + 52
                                           x 100
A-B-cc    46                6951
aabbC-    52       = 28.4 mu
aaB-cc    990
                                       I
A-bb-C-   887      A C             B
A-bb cc   600
                   a c             b
aaB-C-    589
           #     Region II
A-B-C-    2001
aabbcc    1786   600 + 589 + 46 + 52
                                       x 100
A-B-cc    46             6951
aabbC-    52       = 18.5 mu
aaB-cc    990
                       II
                     18.5 mu    28.4 mu
A-bb-C-   887      A C          B
A-bb cc   600
                   a c          b
aaB-C-    589
Genetics
 In the News
Fig. 4.18. Molecular Markers (RFLP)
           EcoRI cleavage sites
             Molecular Mapping Markers
Fig. 4.19




Fig. 4.20a
Fig. 4.20b
p. 143. Fluorescent dyes are often used to label DNA so that the

    positions of DAN fragments in a gel can be identified.
                  Assignments
• Read from Chapter 3, 3.6 (pp. 100-106),

• Master Problems…3.12, 3.15, 3.20,

• Chapter 4, Problems 1, 2,

• Questions 4.1 - 4.4, 4.6, 4.7, 4.9, 4.11 -4.14, 4.19 -
  4.20 a,b,c,d.

• Exam Friday, Oct. 16th,
   – One hour (you can use the entire 80 minutes, but no
     more). One 8” x 11”, one sided crib sheet.

				
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