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Griffiths AJF, Gelbart WM, Miller JH, et al. Modern Genetic Analysis. New York: W. H. Freeman; 1999.

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Modern Genetic Analysis.

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Independent Assortment

First, because we will now begin analyzing two or more genes in crosses, we need to introduce more symbolism. A useful symbolism widely used by geneticists to represent two genes on different chromosomes is to show them separated by a semicolon. Two examples from diploids are A/a ; B/b and m+/m ; p+/p+. Two examples from haploids are leu+ ; ade and fr+ ; trp+.

We can illustrate the process of independent assortment with a hypothetical example, say, in mice, involving the allele pairs A/a and B/b. We will assume these genes are on different pairs of homologous chromosomes, a short pair and a long pair. First we will cross two pure lines to create a dihybrid, or double heterozygote, in which to study recombination. The two lines we will cross are A / A ; B / B and a / a ; b / b:

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The F1 can be of only one type, as follows:

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Now we can use the law of equal segregation to predict the gametic genotypes produced by the F1, and their proportions. We know that 1/2 the gametes will be A and 1/2 will be a. We also know that 1/2 of the gametes will be B and 1/2 will be b. These two ratios must be multiplied to determine the overall gametic genotypes. Why? Recall that the product law (Chapter 4) states that the probability of independent events is the product of their individual probabilities, so since the two genes in question are acting independently, it is appropriate to multiply the gametic probabilities. The expected gametic frequencies of 1/2 are effectively probabilities of 1/2 (or 50 percent); therefore, the gametic array can be shown graphically as follows:

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Figure 5-2 on the following page shows how this ratio is produced at the level of chromosome movements at meiosis. Basically it is because there are two different but equally frequent ways the spindle fibers can attach to the centromeres:

  • In half the meiocytes A and B are pulled to one pole, and a and b to the other.
  • In the other half of the meiocytes A and b are pulled to one pole, and a and B to the other.

Figure 5-2. Meiosis in a diploid cell of genotype A/a; B/b, showing how the segregation and assort-ment of different chromosome pairs give rise to the 1:1:1:1 gametic ratio.

Figure 5-2

Meiosis in a diploid cell of genotype A/a; B/b, showing how the segregation and assort-ment of different chromosome pairs give rise to the 1:1:1:1 gametic ratio.

The general principle is known as independent assortment of allele pairs. This principle is based on the fact that the equal segregation of one allele pair is independent of the equal segregation of the other allele pair because they are on different chromosomes.


Allele pairs on different chromosome pairs assort independently.

Which of the gametes are recombinants? It is simply a matter of applying the definition of recombination: compare the meiotic input genotypes with the meiotic output genotypes. The input genotypes were the genotypes of the parental gametes that fused to form the F1, and we know they were A ; B and a ; b. The output is the set of gametes produced by meiosis in the F1, as shown in Figure 5-2. The genotypes A ; B and a ; b are the same as the parental input, so these types are not recombinant. However, the gametic genotypes A ; b and a ; B are not the same as the input, so by definition they are recombinants. In summary

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The total recombinant frequency (RF) is 1/4 + 1/4  = 1/2 = 50 percent. This 50 percent RF is always observed for genes on different chromosome pairs; in fact, in a situation in which nothing is known about the location of two genes, observing an RF value of 50 percent in the sex cells from a dihybrid meiocyte will immediately suggest that the genes are on different chromosome pairs. However, loci that are very far apart on the same chromosome can also show an RF value of 50 percent, so more experimentation is needed to be sure of the exact situation.

The analysis will be exactly the same if instead of starting with two pure-breeding strains A/A ; B/B and a/a ; b/b, we start with A/A ; b/b and a/a ; B/B. Here the parental gametes are A ; b and a ; B, so the F1 is A/a ; B/b as before, and because of the law of segregation, there will be the same types of gametes in the same proportions:

Image ch5e4.jpg

Now the recombinants will be A ; B and a ; b, but they will still be at a frequency of 50 percent.


In a dihybrid A/a ; B/b involving two genes on two different pairs of chromosomes, the recombinant frequency will always be 50 percent.

So far we have not addressed the problem of how to identify the genotypes of the products of meiosis issuing from a dihybrid meiocyte. In haploid organisms this is simple (as we saw in the Neurospora example above) because the products are the sexual spores (such as ascospores in fungi), and these can easily be grown up and genotypes assigned directly. However, in a diploid organism such as an animal the gamete genotypes cannot be observed directly, so the individual under investigation must be crossed, to test the gametes indirectly. There are two types of crosses that can be used to measure recombination in a diploid, a testcross and a self.

Testcross of a Dihybrid

The best way to determine the genotypes of the gametes of a dihybrid diploid is to make a cross to a tester, an individual that carries only recessive alleles for the genes under investigation. Such a cross is called a testcross. A tester must be fully homozygous recessive, for example a/a ; b/b. Since the gametes of the tester carry only recessive alleles, the genotypes of the gametes of the dihybrid will be expressed in the phenotypes of the testcross progeny. The general nature of the testcross is illustrated in Figure 5-3 on the following page. If the genes are on different chromosomes, independent assortment will produce a 1:1:1:1 ratio of gametes, which in turn will produce a 1:1:1:1 phenotypic ratio in the progeny:

Image ch5e5.jpg

Figure 5-3. The detection of recombination in diploid organisms.

Figure 5-3

The detection of recombination in diploid organisms. Recombinant products of a diploid meiosis are most readily detected in a cross of a heterozygote to a recessive tester.

The chromosomal assortment that leads to this result is summarized in Figure 5-4.

Figure 5-4. Independent assortment, which always produces a recombinant frequency of 50 percent.

Figure 5-4

Independent assortment, which always produces a recombinant frequency of 50 percent. This diagram shows two chromosome pairs of a diploid organism with A and a on one pair and B and b on the other. Note that we could represent the haploid situation by (more...)

In these examples we have illustrated independent assortment with genes known to be on different chromosomes. However, the standard ratios such as the 1:1:1:1 testcross progeny ratio can be used to infer that genes are assorting independently. We can illustrate this with a cross used by Mendel. Recall from Chapter 4 that Mendel deduced that the two pea color phenotypes yellow and green were determined by two alleles of one gene, Y (yellow) and y (green). Through other analyses he knew that two pea shape phenotypes, round and wrinkled, were determined by alleles of another gene, with R determining round and r wrinkled. Since in this case we don’t know if the two genes are on separate chromosomes or on the same chromosome, we must devise a way of representing this uncertainty symbolically. There is no universally agreed- upon convention for this situation, but in this book we will use a period or dot to show this uncertainty. From his cross of a pure-breeding round, yellow plant (R/R.Y/Y) with a pure-breeding wrinkled, green one (r/r.y/y) Mendel obtained a dihybrid F1 that was (as expected) round, yellow (R/r.Y/y). When these F1 individuals were testcrossed to an r/r.y/y tester, the following progeny were produced:

Image ch5e6.jpg

Mendel saw that these numbers were very close to a 1/4:1/4 :1/4:1/4 ratio, so he deduced that the two pairs of al-leles were assorting independently, and we now know this is because they are located on different chromosome pairs. We would now rewrite the genotype of the dihybrid as R/r ; Y/y.

Self of a Dihybrid

Sometimes a tester is not available; however, independent assortment can be demonstrated in a self of a dihybrid (or a cross of identical dihybrids). As an example we can use a dihybrid system from Gregor Mendel’s work. The cross was between two dihybrid parents

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We know that if the genes are on different chromosome pairs, the gametes of both parents will be in the ratio

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The fusion of the male and female gametes can be represented by a 4 × 4 grid as shown in Figure 5-5.

Figure 5-5. Punnett square showing predicted genotypic and phenotypic constitution of the F2 generation from a dihybrid cross.

Figure 5-5

Punnett square showing predicted genotypic and phenotypic constitution of the F2 generation from a dihybrid cross.

Each of the 16 squares in the grid represents 1/16 of the total progeny outcomes, so the overall phenotypic ratio in the progeny is

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This 9:3:3:1 ratio is typical of selfed dihybrids showing independent assortment. In general the progeny ratio of a self of a dihybrid for genes on separate chromosomes can be written:

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A photograph of a 9:3:3:1 phenotypic ratio in corn (Zea mays) is shown in Figure 5-6.


A 1:1:1:1 ratio in a testcross of a dihybrid and a 9:3:3:1 ratio in a self of a dihybrid both reflect a gametic ratio of 1:1:1:1, which shows the allele pairs are assorting independently (generally because they are on different chromosome pairs) and that the RF is 50 percent.

Figure 5-6. A 9:3:3:1 ratio in the phenotypes of kernels of corn.

Figure 5-6

A 9:3:3:1 ratio in the phenotypes of kernels of corn. Each kernel represents a progeny individual. The progeny result from a self on an individual of genotype A/a ; B/b, where A = purple, a = yellow, B = smooth, (more...)

Calculating Phenotypic and Genotypic Ratios for Independently Assorting Genes

In Chapter 4 we calculated the phenotypic ratios expected from progeny of crosses involving alleles of single genes. For example

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This knowledge makes it possible to predict genotypes and phenotypes involving several genes if they are assorting independently. For example in the cross A / a ; b / b  × A / a ; B / b, we might want to calculate the expected phenotypic ratio in the progeny. We know the progeny ratio for the first gene will be 3/4 A / –: 1/4 a / a, and for the second it will be 1/2 B / b: 1/2 b / b. Therefore, if these genes assort independently, we can combine the two phenotypic ratios randomly by drawing a device called a branch diagram. In the branch diagram the progeny ratios are derived simply by multiplying the proportions according to the product law, as shown below:

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The same progeny ratio could be derived using a grid, which also portrays random associations. Note that the axes of this grid show phenotypic proportions. (Grids are very useful in genetics and can be used in numerous ways.)

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In a more complex cross, say, A/a ; B/b ; C/c × a/a ; B/b ; C/c, we might want to predict the proportion of some specific progeny genotype required for an experiment, perhaps the genotype a/a ; b/b ; c/c for use as a tester. Since we know the genotypic proportions for individual genes will be

Image ch5e14.jpg

If the genes assort independently, we can use the product rule simply to multiply all the proportions of the desired single gene genotypes (shown in bold) to obtain the expected proportion (probability) of a/a ; b/b ; c/c; it will be 1/2  × 1/4  × 1/4  = 1/32 . Therefore if we need this genotype, we would have to screen more than 32 progeny to stand a reasonable chance of obtaining one that is a/a ; b/b ; c/c.

These simple calculations based on independence can be used to predict phenotypic, genotypic, or gametic proportions in crosses if it is known or assumed that the genes are assorting independently. However, as we saw in Chapter 4 and above, genetic analysis works in two directions, so we can also use specific progeny proportions to make deductions about the genotypes and phenotypes of the parents, if these are not known. For example, what could we deduce if we were to self some plant with the normal wild-type phenotype of blue, large petals and observed the following numbers of phenotypes in the progeny:

Image ch5e15.jpg

We would note that the progeny numbers represent a very close fit to a 9:3:3:1 ratio, so we could deduce that


The parent must have been dihybrid for two genes affecting petal color and petal size.


Blue petal is dominant to white, and large petal is dominant to small. We could invent allele symbols w+ = blue and w = white, and s+ = large and s = small.


The two genes are assorting independently and are most likely on different chromosome pairs; the selfed plant must have been of genotype w+/w ; s+/s.

By agreement with the publisher, this book is accessible by the search feature, but cannot be browsed.

Copyright © 1999, W. H. Freeman and Company.
Bookshelf ID: NBK21386


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