Mendelian Genetics: Linkage

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Before continuing with this page you might want to review independent assortment and dihybrid crosses.

Linkage.

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Linkage refers to a situation where two loci are on the same chromosome. Recall that if genes at two loci are on separate chromosomes, then they segregate independently. These loci are said to be uninked.  This figure shows the tetrad during prophase 1 before crossing over takes place in a heterozygote with genotype AaBb, then shows one of many cross over events that could lead to new recombinants. Maternal chromosomes are shown in blue and paternal in red.

The next panel in this figure shows a cross over event that can lead to new chromosome types.  These sorts of crossovers occur with frequency r where r is the frequency of recombination, while the third panel shows the four possible chromosomes after meiosis. and their frequencies. Generally the two chromosomes (1,4 ) are most common. These are often called the non recombinant or parental chromsomes because there has been no detectable crossing over between the loci during meiosis. This is because when sister chromatids swap segments, no recombination happens sunce sister chromatids start out prophase I with identical DNA. Chromosomes 2, 3 are recombinants which happen because of crossing over during meiosis.

Frequency of recombination

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The frequency of recombination, r  between two loc is between 0 and 1/2, depending on how closely linked the loc are to each other on the chromsome. If the locus "A" and locus "B" are right next to each other then r will be very small. This is because where the cross over event takes place along the chromatids is random. Thus when the loci are close together, they tend to segragate together.

If the loci are far apart, say near opposite ends of the chromosome, then r may approach 1/2.  Imagine the limiting situation where r = 1/2. Then the chromsomes resulting from meiosis are equally frequent. This means that the gametes with genotypes AB, Ab, aB and ab are equally frequent just as they are if the loci are unlinked.  Thus r is a measure of the degree of independent assortment between loci which is introduced by crossing over.  If only we had a way to detect the different chromsome types in this situation then we might have a way of  figuring out where the loci of different genes are.  It turns out that we do and the method is based on the test cross for independent assortment.

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Importance of  linkage

We are interested in linkage for two reasons:

Two point test cross for detecting linkage 

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Test cross results under independent assortment

Recall that a dihybrid cross for two Mendelian traits with complete dominance gives a 9: 3 : 3 :1 phenotypic ratio assuming indpependent assortment.  Consider the following cross:  AaBb X aabb.  You should be able to verify, using branch diagrams or a Punnett square that the ratio of phenotypes for the offspring from this cross is 1 AB  : 1 Ab : 1 aB : 1 ab.  Here phenotypes are represented by italicized letters.  Note that the phenotypes from this cross just happen to correspond to the genotypes of the gametes from the heterozygote parents.  In situations where there is linkage we can thus literally read the chromosome types  if we do the proper test cross.

Procedure for the two point test cross.

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The procedure for doing this test cross is to begin with parents which  pure breeding for phenotypes  AB and for ab. These of course have corresponding genotypes AABB and aabb.  If there is linkage we have the added bonus that these parents have the following chromosome arrangement:

AB/AB  and ab/ab respectively. Here the letters on each side of the slash represent chromosome arrangements.  When these parents mate the F1's have the following chromsome arrangement:  AB/ab. This arrangement corresponds to the top panel in my figure shwoing the results for crossing over in one of these F1 individuals when it produces gametes.  The four possible gametes then are AB , Ab aB ab correspnding  to the chromsomes in the bottom panel of the figure. The colors are meant to help you see the relationship between this notation and what is happening in my figure.

Thus when we take the F1's and cross them with an indivudal with genotype aabb, we can then essentially read off the chromsome types from the genotypes we observe from the results of this test cross. This sort of test cross is called a two point test cross because we are using it to gauge the degree of linkage between two loci or 'points' along the chromosome.

Test cross results under linkage

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Chi square analysis for linkage

Note: If you are not familiar with the concepts of statistics and hypothesis testing, you might also want to see this discussion on hypothesis testing in genetics.

Consider the following example examined by Thomas Hunt Morgan and discussed in Griffiths et al 1993. He investigated linkage between the gene for purple eye color(pr) and the gene for vestigial wing. He first mated flies known to be pure breeding for purple eye and vestigial wing with wild type flies pure breeding for both traits. Let pr be the allele for purple eye and vg be the allele for vestigial. Both these alleles are recessive. For clarity, let pr+ and vg+ be the corresponding wild type alleles. The F1s all are of genotype pr pr+ vg vg+, written in our chromosomal notation pr vg /+ + .

When Morgan mated female offspring from the test cross with male pr vg / pr vg  flies, he obtained the following results:

Phenotype


Genotype

Number observed Oi

Number expected Ei

(Oi -Ei)^2/E

Wild type eye,
Wild type wing

pr+ vg+ / pr vg

1139

659.75

348.132721

Wild type eye,
Vestigial (vg) wing

pr+ vg / pr vg

151

659.75

392.310061

Purple eye(pr),

Wild type wing
pr vg+ / pr vg

154

659.75

387.69695

Purple eye(pr),Vestigial (vg) wing


pr vg / pr vg

1195

659.75

659.75

total

2639

2639

1787.88973

The observed Chi square, 1787.9, is clearly much better than the critical chi square (df = 7.81) at the 0.05 level of significance. Thus the hypothesis that the genes are unlinked is rejected. In this case the results are very likely to be related to linkage. However, one must be very careful to check that the hypothesis being tested is appropriate for a use of a simple chi square. Deviations from the expected 1 : 1 : 1 : 1 ratio may also arise if zygotes from a particular female mating with the wild type male differ in viability. Further even if linkage exists, viability effects between the cross over gametes will inflate the Chi square test leading to a greater chance of rejecting the hypothesis of linkage.

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Map distance

The map distance between any two loci is defined in terms of map units, where w\one map unit(m.u.) is equal to 1% recombination.  The map distance between two loci then is given by the following formula:

(Number of recombinants observed/total offspring scored ) X 100

For Morgan's data the map distance between the loci for pruple eye color and vestigial wing is:

(151+ 154)/2639 x 100 =  11.6 map units

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Map Distance and physical distance between loci

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Map distance is terms of map units is not the same as physical distance between loci. If loci are far apart map distance tends to under estimate the degree of physical distance between the loci. This is because when loci are far apart, double cross overs between the loci become more common. If a single cross over causes recominant chromatidss Ab and aB, cross over between the A and B loci of thes chromatids leads to chromatids AB, and ab again. This will become more clear in the following section.

Three point test cross and chromosome mapping

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Linkage between 3 gene pairs can be analyzed with an extension of the two point test cross. Three other fruit fly mutations-cinnabar eye, curved wing and plexus(extra wing veins) are known to be on chromosome 2. A geneticist did a cross using pure breeding wild type individuals for all three characteristics mated with a pure breeding wild type flies for all three traits. Using cn for the cinnabar mutant allele, c for the curved wing mutant  allele and px for the plexus wing mutant allele, the geneticist obtained the following data:

Phenotype

Genotype

cn, c, px

cn c px / cn c px

296

cn, c, +
cn c + / cn c px

63

cn, +, + 


cn + + / cn c px

119

cn, +, px


cn + px / cn c px

10

+, c, px


+ c px / cn c px

86

+, c, +


+ c + / cn c px

15

+, +, px


+ + px / cn c px

82

+, +, +


+ + + / cn c px

329

Total

1000

 

The basic strategy for anlyzing this sort of three point test cross data is to tally the number of recombinant and non recombinat chromsomes scored for each pair of loci. The we use this data to calculate the number of map units between each pair of loci. So for this example we have the following three pairs of loci and number of recombinat and non recombinant chromsomes:

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Loci
Recombinants
Non-recombinants
Total
Map distance in Map units
cn,c
119 +10 + 86 + 15 = 230
296+63 + 329 + 82 = 770
1000
23
cn, px
63 + 119 + 86 + 82 = 350
296 + 10 + 15 + 329 = 650
1000
35
c,px
63 + 15 + 82 + 10 = 170
296 + 86 + 329 + 119 = 830
1000
17

Consider the recombinants involved in the loci cn, c. These are going to be chromosomes cn + and +cn so the number of combinations with these genotypes regardless of what happens at the other louc are added together. This gives 119 +10 + 86 + 15 = 230  as shown in the upper left hand set of numbers in the table. Note that one could subtract this number from the total offspring counted (1000) to get the total of non-recombinants, but it is a good idea to tally these separately as a check on arithmetic as I have done in my table.

The map units separating the loci in this example can be read off from the table and are shown in the last column. The data suggests the following arrangement  of  loci. Since the cn, px loci are farthest apart (35 m.u),  the c locus must be in the middle. Arbitrarily putting the cn locus first we have the following map.

  <-----------------35----------->                                             
cn-----------------c-------------px
            23                    17

Notice that the distances are not additive. 23 + 17 = 40, not 35 as expeccted. This again is because when loci are separated by large distances, additional cross overs flip the gene arrangement back to the non recombinant state. These extra cross overs cannot be detected directly by the 3 point cross. Thus when geneticsists map the location of loci on a chromosome, they try to find loci that are close togther, and string the map unit distances together to get a more accurate estimate for longer distances.

                               

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