1. Gene Mapping in Eukaryotes
Text authored by Dr. Peter J. Russell
Slides authored by Dr. James R. Jabbur
CHAPTER 14
Gene Mapping in Eukaryotes

2. Introduction Genes on nonhomologous chromosomes assort independently
Syntenic genes may be inherited together and belong to a linkage group
Classical genetics analyzes the frequency of allele recombination in the progeny of genetic crosses
New associations of parental alleles are recombinants, produced by genetic recombination
Testcrosses help to determine which genes are linked and a linkage map (genetic map) is constructed for each chromosome

3. Morgan’s Experiments with Drosophila
Both the white eye gene and the gene for miniature wings are on the fly X chromosome
Female white eyed, miniature winged flies were crossed with wild-type males (wm/wm x w+m+/Y)
In the F1, all males were white-eyed with miniature wings (wm/Y) and all females were wild type for both eye color and wing size (w+m+/wm)
F1 interbreeding was then undertaken (this action is the equivalent of a testcross, as we know the genotypes)
In the F2, the most frequent phenotypes for both sexes were the phenotypes of the parents in the original cross (white eyes with miniature wings, and red eyes with normal wings)
Nonparental phenotypes (white eyes with normal wings or red eyes with miniature wings) occurred in about 37% of the F2 flies. This is statistically well below the 50% prediction rate for independent assortment, indicating that non-parental flies result from the recombination of linked genes

4. Figure 15.1 Morgan’s experimental crosses of white-eye and miniature-wing variants of Drosophila melanogaster, showing evidence of linkage and recombination in the X chromosome.

5. From his studies, Morgan proposed that:
During meiosis, alleles of some genes assort together because they are near each other on the same chromosome
Recombination occurs when genes are exchanged between the X chromosomes of the F1 females (it must occur in the female population because males do not have 2 homologous X chromosomes)
Repeated studies produced similar results

6. Recombination and Exchange
Proof that physical exchange between chromosomes results in genetic recombination came with studies in Drosophila
Genetic and physical (cytological) markers were employed to analyze genetic recombination in meiosis
Curt Stern worked with 2 linked gene loci, car and bar (B):
The car eye color gene is recessive; red eyes are wild-type
The bar (B) eye shape gene is incompletely dominant
Wild-type eye shape is round
Heterozygote eye shape is kidney-shaped
Recessive eye shape is bar-shaped
The crosses and results of allelic segregation are shown
Cytological examination confirmed that physical crossing-over results in genetic recombination
Animation: Recombination & Exchange

7. Constructing Genetic Maps
Detecting Linkage through Testcrosses
Linked genes are used for mapping. They are found by looking for deviation from the frequencies expected from independent assortment
Employing a testcross where one parent is true-bred, homozygous recessive works well for analyzing linkage:
If the alleles are not linked and the second parent is heterozygous, all four possible combinations of traits will be present in equal numbers in the progeny
A significant deviation in the aforementioned ratio (more parental and fewer recombinant types) indicates linkage

8. Animation: Chi-Square Test
Chi-square analysis is used to analyze the testcross data and determine whether a deviation is “significant.” A null hypothesis (“the genes are not linked”) is used because it is not possible to predict phenotype frequencies produced by linked genes
In this case, the deviation is significant and the hypothesis is not valid (thus, the genes are linked)
Animation: Chi-Square Test

9. 1479.4
A P value of less than 5% (0.05) indicates a poor fit, results are rejected

10. The Concept of a Genetic Map
In an individual heterozygous at two loci, there are two arrangements of alleles:
The cis (coupling) arrangement has both wild-type alleles on one homologous chromosome, and both mutants on the other (i.e., w+m+ and wm)
The trans (repulsion) arrangement has one mutant and one wild type on each homolog (e.g., w+m and wm+)
A crossover between homologs in the cis arrangement results in a homologous pair with the trans arrangement. A crossover between homologs in the trans arrangement results in cis homologs.

11. Crosses in Drosophila showed that crossover frequency for linked genes is characteristic for each gene pair (the frequency stays the same, whether the genes are arranged in cis or trans)
Sturtevant (Morgan’s student) used recombination frequencies to make a genetic map (a 1% crossover rate is a genetic distance of 1 map unit or centiMorgan)
The first genetic map was based on crosses involving 3 sex-linked genes (w, white eyes; m, miniature wings; y, yellow body color)

12. …important point
The farther apart the two genes are on the chromosome, the more likely it is that a crossover will occur between them and therefore, the greater the occurrence of their crossover (frequency)

13. Gene Mapping with a 2 Point Testcross
With autosomal recessive alleles, when a double heterozygote is crossed, four phenotypic classes are expected
If the genes are linked, the two parental phenotypes will be about equally frequent and more abundant than the two recombinant phenotypes

14. Generating a Genetic Map
A genetic map is generated from estimating the crossover rate in a particular segment of the chromosome
If you have linked genes with a crossover:
R.F. (mu) = # observed recombinants X 100
# observed total progeny
Problem: It may not exactly match the physical map:
The nonrandom distribution of crossing-over limits the accuracy of mapping. Crossover events are not equally probable at all sites on the chromosome (hotspots exist)***
Double crossovers between two loci can restore the partental genotype (as will any number of even crossovers) (next slide)

15. Here:

16. Gene Mapping with 3 Point Testcrosses
Typically, geneticists design experiments to gather data on several traits in one testcross
In the progeny, each gene has two possible phenotypes. For 3 genes, there are 8 expected (23) phenotypic classes in the progeny
Establishing the order of genes on the chromosome is determined from the phenotypic results of the cross
Two classes will be parental (and most abundant)
Of the six remaining classes, 2 will be present at the lowest frequency, resulting from an apparent double crossover, thereby establishing the apparent gene order
Animation: 3 Point Mapping

17. The above figure demonstrates the consequences of a double crossover in a triple heterozygote for three linked genes
In a double crossover, the middle allelic pair changes its orientation relative to the outside allelic pairs, producing 2 parental and 2 recombinant (allele) gametes

18. The apparent gene order is determined by the progeny which accounts for the lowest apparent frequency
This group represents the double recombinant
Thus, in the example shown, the correct gene order is rearranged to pjr (look at the genotype in the table for the clue)

19. With the gene order properly arranged, we can calculate the recombination frequencies for the genes
RF = sco + dco x 100%
total progeny
The percentage of progeny generated by crossing-over between p and j is 20.8% ( /500); between j and r is 10% ( /500)
To compute the map distance between outside genes, simply add the two map distances

20. Of course, the ultimate means of generating a genetic map is high throughput sequencing

21. What Progeny Looks like? What kind of crossover took place?
Filial Phenotype
What Progeny Looks like?
What kind of crossover took place?
Number of Observed
ABC
First parent
None
401
Abc
Second parent
389
ABc
Recombinant types
Double (around C) 4
abC 6
AbC
Single (between C and B) 75
aBc 65
aBC
Single (between C and A) 35 25
1000

22. What Progeny Looks Like?
Filial Phenotype
What Progeny Looks Like?
Which traits were inherited from one of the parents?
Which trait came from the other parent?
ABC
First Parent
None
abc
Abc
ABc
Recombinant Types
AB from the ABC Parent c
abC
ab from the abc parent C
AbC
AC from the ABC parent b
aBc
ac from the abc parent B
aBC
BC from the ABC parent a
bc from the abc parent A

23. Calculating Recombination Frequency
Map distance between A and C:
All double crossovers + all single crossovers involving A and C
total
Map distance between C and B:
All double crossovers + all single crossovers involving B and C