Independent assortment, linked genes, and recombination

Updated: Dec 30, 2019

Mendel proposed the Law of Independent Assortment to explain his observations that the outcomes for one gene did not impact the outcomes for another gene. When he did crosses for multiple traits, new combinations occurred in the F2 generation that were not present in the P generation. For example, crossing true breeding pea plants for yellow round seeds and green wrinkled seeds can result in an F2 generation that includes both yellow wrinkled seeds and green smooth seeds.

Dihybrid cross showing independent assortment

If genes are located on different chromosomes, a heterozygote has an equal chance of the dominant or recessive copy ending up in each gamete. This means that in a dihybrid cross there are four combinations that are equally likely to be passed on to the next generation.

Allele combinations produced by a heterozygote parent

To find the expected outcomes for crossing two heterozygotes for both traits, you can make a sixteen box Punnett square. Each parent has four different combinations that can be passed on that will be positions along the top and side of the square. You can also skip the larger Punnett square and use math to find the combined probability. Multiply the chance of each trait occurring to get the probability that both traits appear together in the offspring.

Punnett squares and combined probability calculations for a dihybrid cross

I prefer this method because it can be scaled up to three or more traits without becoming more difficult. For example, the chances of an individual having the dominant phenotype of three independently assorting traits is 27/64 (3/4 x 3/4 x 3/4). While you could make a 64 box Punnett square, the more boxes there are, the more chances of making a mistake.


Of course, all of this only works if the genes are unlinked. Alleles of genes close to each other on the same chromosome don't have an equal chance of being sorted into different gametes. Through crossing over, linked alleles may end up in different gametes, but at a lower rate than if they were on different chromosomes (or far apart on the same chromosome). The further apart linked gene are located on the chromosome, the more likely they are to be separated through crossing over. When this happens, the chromosome is called a recombinant. Offspring that show a phenotype combination that resulted from crossing over are also called recombinants.


Let's walk through an example of what will happen with linked genes when we cross a parent with both dominant traits to a parent with both recessive traits, using the fictional scenario in the Heredity IV simulation. In this case, the lack of pinna (external ear) and black eyes are dominant to the presence of pinna and white eyes. When our P cross is between black eye/no pinna and white eye/pinna parents (the genes are not sex linked, so it doesn't matter if males or females have the dominant or recessive traits) the F1 generation results are 100% heterozygous black eye/no pinna. Because the P generation parents are homozygous this is the same result we get when doing a dihybrid cross for unlinked genes (see Heredity I simulation).


P generation cross

When we cross the F1 generation, though, the results will not be the 9:3:3:1 ratio that we expect from independently assorting genes.

Black eye/no pinna x white eye/pinna P generation and results from F1 cross.

If no crossing over occurs during gamete formation in this F1 cross, 75% of the resulting offspring will have both dominant phenotypes and 25% will have both recessive phenotypes. When crossing over occurs, some gametes will have a different combination of alleles on the chromosome and may lead to offspring with one dominant and one recessive phenotype (black eye/pinna and white eyes/no pinna). These offspring are the recombinants. A lower than expected (3/16 + 3/16 = 3/8 or 37.5%) number of recombinant offspring indicates that genes are linked. In the sample data in the picture above, only 2% of the offspring are recombinants. However, in this case, it is difficult to estimate the rate of recombination because crossing over also results in offspring with the dominant phenotype (black eye/no pinna).

Possible chromosome combinations of F1 cross (P generation: black eye/no pinna x white eye/pinna)

In an independently assorting dihybrid cross, a P generation in which one parent has the dominant version and the recessive version of the other and the other parent has the reverse, the F2 generation will have the same 9:3:3:1 expected results. When the two gene are linked, however, the results are different than what occurred when one parent was dominant for both traits and the other parent was recessive for both.

This example uses a P generation cross of black eye/pinna x white eye/no pinna. While the F1 is still 100% dominant phenotypes (black eye/no pinna), each chromosome is what would have been classified as recombinants in the last example.

Possible chromosome combinations of F1 cross (P generation: black eye/pinna x white eye/no pinna)

In this case, the rarest F2 phenotype will be both recessive traits (white eye/no pinna). In the sample data below, only 1% of the offspring have white eyes and the presence of pinna (both recessive traits). Because crossing over can lead to a phenotype that is also possible without crossing over (black eye/no pinna), this cross isn't ideal for estimating the rate of crossing over.

Black eye/pinna x white eye/no pinna P generation and results from F1 cross.

In order to estimate the recombination rate between the genes we need a cross where there isn't overlap in the phenotypes resulting from crossing over and no crossing over. This can be done with a backcross. In a backcross, F1 heterozygotes are crossed to the parent with recessive phenotypes.

Expected results of un-linked backcross

If we perform a backcross with unlinked genes, we should get a 1:1:1:1 ratio (see picture above, using pea seed traits). If the genes are linked there will be an over representation of the parental phenotypes (dominant for both traits and recessive for both traits).

Backcross chromosome combinations

In this case, unlike the F1 cross examples discussed previously, a crossing over event between the two genes will lead to a recombinant phenotype (black eye/pinna and white eye/no pinna). For this reason, the backcross gives us a way of determining the rate of recombination, which is indicative of the distance between the genes.

Backcross results

The % recombination is equivalent to the map units/centiMorgans (cM) between the genes. This is a relative measure of distance on chromosomes and can be used to map the locations of genes. In the sample data in the picture above, 6.6% of the offspring are recombinants. This data suggests that the eye color and pinna genes are 6.6 cM apart. However, remember that it is best practice to base conclusions on an average of multiple trials.


If genes are located far enough apart on a chromosome, crossing over will occur often enough that the genes assort independently. At 50 cM the recombination rate is 50%, which means the expected 1:1:1:1 ratio for unlinked genes is observed.


For more information on the discovery of linked genes and construction of the first fruit fly chromosome linkage maps, see this article from Nature Education.



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