Lecture 5. Linkage, Recombination and the Mapping of Genes on Chromosomes

 

I. Importance

Reconciled exception of independent assortment with mendelian genetics

Establishing linkage of a trait with markers is the FIRST step in the full genetic description of a gene

 

II History

Bateson and Punnett, deviation from 9331 ratios in sweet peas

F1 produced far more parental type of gametes than should have been produced by independent assortment

 

III. Morgan and Colleagues

X-chromosome linkage

Homologous chromosomes occasionally exchange parts in a process called crossing-over

 

The new combinations are called crossover products

 The crossover corresponds to chiasmata

 

Used the chromosome theory of gene location to provide a physical explanation of the genetic results

 

Linkage: when 2 loci are on the same chromosome

 

 

Meiotic recombination: any meiotic process that generates a haploid product with a genotype that

 differs from both haploid genotypes that make-up the meiotic diploid cell. Recombination occurs by

 independent assortment AND by crossing-over.

 

Linkage Maps: Why does the amount of recombinant gamete types or strength of linkage vary so much

 from locus to locus???

 

 

Sturtevant: The amount of recombination is proportional to the distance on a genetic map and

  roughly corresponds to the distance between loci on a chromosome.

 

Recombination frequency (RF) can be used to mark distance apart genes are from one another

1% RF = 1 centimorgan or 1 map unit (cM, m.u.)

Three-point cross

observed:

 

 0.123 x 0.064 = 0.0079 = 0.79%

 

(13 + 9)/4197 = 0.0052 = 0.52%

 

Chromosomal interference- the occurrence of one crossover lowers the chance of another crossover

   occurring nearby

 

Coefficient of coincidence = Frequency observed/Frequency expected;

0.52/0.79 = 0.66

 

Interference = 1 – Coeff. Coincidence = 0.34 and varies from 0-1

 

IV. Evidence for Recombination

 

V. Quantitative Tests for Linkage- Chi Square

 

VI. Tetrad Analysis

Fungal life cycles

Unordered Tetrad Analysis

When PD = NPD then loci are unlinked;

PD>NPD then loci are linked

 

Parental haploid cells:   arg3 ura2  x  ARG3 URA2

 

F1 Diploid cell:  arg3 ura2 / Arg3 URA2

 

                                    Meiosis

 

PD                               NPD                      T

 

 arg3  ura2                                arg3 URA2        arg3 ura2

 arg3  ura2                    arg3 URA2        arg3 URA2

 ARG3 URA2               ARG3  ura2       ARG3 URA2

 ARG3 URA2               ARG3 ura2        ARG3 ura2

 

127                                           3                         70

 

 

All spores in an NPD are recombinant; ½ spores in a Tetratype are recombinant

 

 

RF = (NPD +1/2(T))/Total tetrads  x 100

= (3 + 1/2x70)/200   x 100  = 19 m.u.

 

 

 

Same result if you examine each individual spore:

 

200 tetrads, of which each the 3 NPD’s have 4 recombinants and each of the 70 T have 2 recombinants or

 

 (12 + 140)/800  x 100 = 19 m.u.

 

Tetrad analysis confirms that recombination occurs at the 4-strand stage

 

Only ¼ of all rarer double cross-overs = NPD

 

T >> NPD when there is linkage

 

 

Thus recombination occurs after the chromosomes have replicated, when there are 4-chromatids for each pair

 of homologues.

 

Since recombination occurs after duplication then upper limit of 50% Recombination for genes on same

 chromosome:

 

            single x-over = 50%

            ½ double x-over generate Non parental combinations and ½ parental combinations

 

Tetrad analysis reveals that recombination

 is USUALLY reciprocal

 

  no new alleles are formed?

 

  both classes of recombinants are found in equal frequency almost all of the time- more later

 

 

 

Ordered Tetrad Analysis

 

Numerical example of Ordered-Tetrad Analysis          

 

Tetrad

Group               A            B          C         D         E            F           G

 

 

Seg.                   t  a     t   a       t   a      t  a+      t  a+    t  a+                 t  a

Pattern             t  a       t+ a       t  a+     t+ a      t+ a       t  a+                  t+ a+

                                    t+ a+     t+ a+   t+ a      t+ a+     t+ a      t+ a                  t+ a+

                        t+ a+                t  a+      t+ a+    t  a       t  a+     t+ a      t a

 

Total                 72        16         11            2                      2                     1            1

(105)

 

 

Centromere-t distance: (16 + 2 + 2 + 1)/105  x 100 = 10 mu

 

 

Centromere-a distance: (11 + 2 + 2 +1)/105 x 100 = 7.6 mu

 

 

Are t and a linked? PD (72 + 1) >>NPD (2+1)

 

 

RF = (NPD + 1/2T)/Total  x 100

     = (2 + 1 + ½(16 + 11 + 2))/105 x 100 = 16.7 m.u.

 

 

VII. Conclusion

 

Because genes are on chromosomes they can be physically linked to one another.

 

The amount of recombination between genes (loci) can be used to quantify the distance between genes.

 

 

VIII. Terms/Concepts to know:

 

Recombination, homologous chromosome, recombination frequency, centimorgans, map unit,

chromosomal interference, coefficient of coincidence,interference, Chi-square,

ordered and unordered tetrad analysis, parental ditype, nonparental ditype, tetratype,

 HOW TO MAP, 3-POINT CROSS!! Table 5.1, Figs 2,3,7,11,12,13,15,16,17,19,10,22,23