Linkage Notes

Every individual organism bears several heritable characters. Which are represented by the innumerable genes present on the chromosomes. During meiosis, the chromosomes move into the gametes as units, all the genes present on any given chromosome will segregate as a group and move together from generation to generation. This tendency of the genes located on the same chromosome, to stay together in hereditary transmission, is known as linkage. The genes located on the same chromosome are called linked genes.

All the genes located on a particular chromosome, form a linkage group. Since the genes present on a particular chromosome have their alleles located on its homologous chromosome, genes on a pair of homologous chromosomes. Hence, the number of linkage groups corresponds to the number of haploid chromosomes found in a species.

Drosophila melanogaster has four linkage groups which can be distinguished into three large and one small linkage groups corresponding to the four pairs of chromosomes. Twenty-three linkage groups are present in humans corresponding to 23 pairs of chromosomes. Pea plant has seven linkage groups, corresponding to the seven pairs of chromosomes.

Complete Linkage

The genes closely located in the chromosome show complete linkage as they have no chance of separating by crossing over and are always transmitted together to the same gamete and the same offspring. Thus, the parental combination of traits is inherited as such by the young one.

Incomplete Linkage

The genes distantly located in the chromosome show incomplete linkage because they have a chance of separation by crossing over and of going into different gametes and offspring.

The principle of linkage was discovered by Bateson and Punnet in 1906 in the sweat pea, plant, Lathyrus odoratus. However, linkage, as a concept was put forth by Thomas Hunt Morgan in 1910 based on his experiment on Drosophila melanogaster.

It was not long from the time that Mendel's work was rediscovered that new anomalous ratio began appearing. One such experiment was performed by Bateson and Punnett with sweet peas. They performed a typical dihybrid cross between one pure line with purple flowers and long pollen grains and a second pure line with red flowers and round pollen grains. Because they new that purple flowers and long pollen grains were both dominant, they expected a typical 9:3:3:1 ratio when the F₁ plants were crossed. The table below shows the ratios that they observed. Specifically, the two parental classes, purple, long and red, round, were over represented in the progeny.

	Observed	Expected
Purple, long (P_L_)	284	215
Purple, round (P_ll)	21	71
Red, long (ppL_)	21	71
Red, round (ppll)	55	24
Total	381	381

At the time of these experiments, Bateson and Punnett were not able to develop an acceptable hypothesis. The best explanation they posed was that in some manner the phenotypic classes (alleles) in the parents were coupled, and they did not sort independently into gametes as predicted by Mendel's second law.

Proof that genes on the same chromosome can at times be inherited as blocks awaited the results of Thomas Hunt Morgan with Drosophila. Morgan crossed red eye, normal wing flies (pr⁺pr⁺ vg⁺vg⁺) with purple eye, vestigal wing (prpr vgvg) flies. The figure below shows the cross and the F₁ genotypes. (The bars are used to shows that the genes reside on the same chromosome.)During meiosis, four different F₁ gametes are produced. The parental gametesare developed without any processing. The recombinant gametes though occur by a process called called crossing over. (The X between the two F₁ chromosomes represents the crossing over event.)

Morgan performed a testcross by crossing prpr vgvg flies to F₁. The testcross is powerful because it allows you to follow the meiotic events in one parent because all of the gametes from the test cross parent are homozygous recessive. For this example, the testcross genotype is pr vg. Therefore the testcross progeny will represent the distribution of the gametes in the F₁. Remember that a testcross to F₁ derived from a dihybrid cross gave a 1:1:1:1 ratio. But this is not what Morgan observed. The following table shows the result of this test cross.

F1 Gamete	Testcross Distribution	Gamete Type
pr+ vg+	1339	Parental
pr+ vg	151	Recombinant
pr vg+	154	Recombinant
pr vg	1195	Parental

These results confirm the Bateson and Punnett hypothesis that two genes do not always assort independently. A further confirmation experiment was performed by Morgan when he crossed red eye, vestigal wing flies and purple eye, normal wing flies. Whereas in the first cross, the two dominant alleles and two recessive alleles were on the same chromosome the F₁, in the is cross a dominant allele was on the same chromosome as a recessive allele. The term for the first chromosomal arrangement of the F₁ is calledcoupling, whereas the second arrangement is called repulsion. Another set of terms to describe these arrangements are cis and trans, respectively. The following shows the chromosomal arrangement for the cross of two parents in repulsion.

As with the first cross, Morgan testcrossed these F₁ flies. The following table shows the distribution of these F₁gametes.

F1 Gamete	Testcross Distribution	Gamete Type
pr+ vg+	157	Recombinant
pr+ vg	965	Parental
pr vg+	1067	Parental
pr vg	146	Recombinant

It was expected that both the coupling and repulsion crosses would yield 1:1:1:1 ratios. How can we determine if the results deviate from this ratio. As with any ratio, we can use the chi-square test to determine if the observed results fit or deviate from the expected ratio. The two tables below show the results for the chi-square for the two crosses.

Recombination and Estimating the Distance Between Genes

Physical crossing over during meiosis I is a normal event. The effect of this event is to rearrange heterozygous homologous chromsomes into new combinations. The term used for crossing over is recombination. Recombination can occur between any two genes on a chromosome, the amount of crossing over is a function of how close the genes are to each other on the chromosome. If two genes are far apart, for example at opposite ends of the chromosome, crossover and non-crossover events will occur in equal frequency. Genes that are closer together undergo fewer crossing over events and non-crossover gametes will exceed than the number of crossover gametes. The figure below shows this concept.

Finally, for two genes are right next to each other on the chromosome crossing over will be a very rare event.
Two types of gametes are possible when following genes on the same chromosomes. If crossing over does not occur, the products are parental gametes. If crossing over occurs, the products are recombinant gametes. The allelic composition of parental and recombinant gametes depends upon whether the original cross involved genes in coupling or repulsion phase. The figure below depicts the gamete composition for linked genes from coupling and repulsion crosses.
It is usually a simple matter to determine which of the gametes are recombinant. These are the gametes that are found in the lowest frequency. This is the direct result of the reduced recombination that occurs between two genes that are located close to each other on the same chromosome. Also by looking at the gametes that are most abundant you will be able to determine if the original cross was a coupling or repulsion phase cross. For a coupling phase cross, the most prevalent gametes will be those with two dominant alleles or those with two recessive alleles. For repulsion phase crosses, gametes containing one dominant and one recessive allele will be most abundant. 

The important question is how many recombinant chromosomes will be produced. If the genes are far apart on the chromosome a cross over will occur every time that pairing occurs and an equal number of parental and recombinant chromosomes will be produced. Test cross data will then generate a 1:1:1:1 ratio. But as two genes are closer and closer on the chromosome, fewer cross over events will occur between them and thus fewer recombinant chromosomes will be derived. We then see a deviation from the expected 1:1:1:1 ratio.

 By definition, one map unit (m.u.) is equal to one percent recombinant phenotypes. In honor of the work performed by Morgan, one m.u. is also called one centimorgan (cM).

Now let's determine the linkage distance between the genespr and vg. We can actually make two estimates because we have the results from coupling and repulsion phases crosses. The coupling phase analyzed a total of 2839 gametes, and of these gametes 305 (151 pr⁺ vg+ 154 pr vg⁺) gametes were recombinant. To determine the linkage distance simply divide the number of recombinant gametes into the total gametes analyzed. So the linkage distance is equal to 10.7 cM [(305/2839)*100)].

We can also perform the same calculations with the results from the repulsion phase cross. For this experiment, a total of 2335 gametes were analyzed, and 303 (151 pr⁺ vg⁺+ 154 pr vg) of these were the result of recombination. The estimate of the linkage distance between pr and vgfrom these experiments is 13.0 cM [(303/2335)*100].
Obviously, we can conclude that the two genes are linked on the same chromosome. But what is the true linkage distance, the 10.7 cM value from the coupling experiment or the 13.0 value from the repulsion experiment? Actually neither is correct or wrong. These again are two estimates. Only by repeating this experiments many times using a number of different independent crosses can we settle on a value.
Once we have settled on a value, these genes can then be graphically displayed. Let's say that the true distance between the pr and vg genes is 11.8 cM, that is the average of our two estimates. We can next display them along a chromosome in the manner shown below. (Note that it is customary to use the allelic designantions of the mutant phenotype when drawing these maps.)

The final point that we need to make regards the maximum distance that we can measure. Because of the way in which the calculations are performed, we can never have more that 50% recombinant gametes. Therefore the maxmimum distance that two genes can be apart and still measure that distance is just less that 50 cM. If two genes are greater than 50 cM apart, then we can not determine if they reside on the same chromosome or are on different chromosomes. In practice though, when experimental error is considered, as distances approach 50 cM it is difficult to determine if two genes are linked on the same chromosome. Therefore, other mapping techniques must be used to determine thelinkage relationship among distantly associated genes. One method that allows us to deal with distantly related genes and to order genes is the three-point cross.

Deriving Linkage Distance and Gene Order From Three-Point Crosses

By adding a third gene, we now have several different types of crossing over products that can be obtained. The following figure shows the different recombinant products that are possible.

Now if we were to perform a testcross with F₁, we would expect a 1:1:1:1:1:1:1:1 ratio. As with the two-point analyzes described above, deviation from this expected ratio indicates that linkage is occurring. The best way to become familiar with the analysis of three-point test cross data is to go through an example. We will use the arbitrary example of genes A, B, and C. We first make a cross between individuals that are AABBCC and aabbcc. Next the F₁ is testcrossed to an individual that is aabbcc. We will use the following data to determine the gene order and linkage distances. As with the two-point data, we will consider the F₁gamete composition

Genotype	Observed	Type of Gamete
ABC	390	Parental
abc	374	Parental
AbC	27	Single-crossover between genesC and B
aBc	30	Single-crossover between genesC and B
ABc	5	Double-crossover
abC	8	Double-crossover
Abc	81	Single-crossover between genesA and C
aBC	85	Single-crossover between genesA and C
Total	1000

The best way to solve these problems is to develop a systematic approach. First, determine which of the the genotypes are the parental gentoypes. The genotypes found most frequently are the parental genotypes.From the table it is clear that the ABC and abc genotypes were the parental genotypes.
Next we need to determine the order of the genes. Once we have determined the parental genotypes, we use that information along with the information obtained from the double-crossover. The double-crossover gametes are always in the lowest frequency. From the table the ABcand abC genotypes are in the lowest frequency. The next important point is that a double-crossover event moves the middle allele from one sister chromatid to the other. This effectively places the non-parental allele of the middle gene onto a chromosome with the parental alleles of the two flanking genes. We can see from the table that the C gene must be in the middle because the recessive c allele is now on the same chromosome as the Aand B alleles, and the dominant C allele is on the same chromosome as the recessive a and b alleles.
Now that we know the gene order is ACB, we can go about determining the linkage distances between A and C, and Cand B. The linkage distance is calculated by dividing the total number of recombinant gametes into the total number of gametes. This is the same approach we used with the two-point analyses that we performed earlier. What is different is that we must now also consider the double-crossover events. For these calculations we include those double-crossovers in the calculations of both interval distances.
So the distance between genes A and C is 17.9 cM [100*((81+85+5+8)/1000)], and the distance between C and B is 7.0 cM [100*((27+30+5+8)/1000)].