Linked genes are inherited together

Some of the crosses Morgan performed with fruit flies yielded phenotypic ratios that were not in accordance with those predicted by Mendel’s law of independent assortment. Morgan crossed Drosophila with known genotypes at two loci, B and Vg:

  1. B (wild-type gray body) is dominant over b (black body).

  2. Vg (wild-type wing) is dominant over vg (vestigial, a very small wing).

Morgan first made an F1 generation by crossing homozygous dominant BBVgVg flies with homozygous recessives (bbvgvg). He then performed a test cross with the F1 flies: BbVgvg × bbvgvg.1 Morgan expected to see four phenotypes in a ratio of 1:1:1:1, but that is not what he observed. The body color gene and the wing size gene were not assorting independently; rather, they were usually inherited together (Figure 12.15).

experiment

image

Figure 12.15A Some Alleles Do Not Assort Independently

Original Paper: Morgan, T. H. 1912. Complete linkage in the second chromosome of Drosophila. Science 36: 719–720.

Morgan’s studies showed that the genes for body color and wing size in Drosophila are linked, so that their alleles do not assort independently.

1Do you recognize this type of cross? It is a test cross for the two gene pairs; see Figure 12.3.

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work with the data

Figure 12.15B Some Alleles Do Not Assort Independently

Original Paper: Morgan, T. H. 1912.

Mendel’s work was “rediscovered” 40 years after its publication. At that time, biologists began to find some exceptions to the rules of inheritance that Mendel had proposed. Thomas Hunt Morgan and his colleagues made dihybrid test crosses in fruit flies. They proposed that the clearest way to test for linkage was not to look at aberrations in the 9:3:3:1 phenotypic ratio expected from an F1 × F1 cross, but to examine aberrations in the 1:1:1:1 ratio expected from an F1 × homozygous recessive test cross (see Figure 12.3). Morgan’s group then hypothesized that linkage had a physical basis, namely that genes are linked together on chromosomes and that rare crossing over during meiosis gives rise to the less frequent phenotypes. Examination of actual chromosomal events confirmed this.

QUESTIONS

Question 1

Morgan first performed a dihybrid cross between black, normal-winged flies (bbVgVg) and gray, vestigial-winged flies (BBvgvg). The F1 flies were interbred, yielding the F2 phenotypes shown in the table (Experiment 1). Compare these data with the expected data in a 9:3:3:1 ratio by using the chi-square test (see Appendix B for information about the chi-square test). Are there differences, and are they significant?

Using a chi-squared test with 3 degrees of freedom the chi-squared value is 429.96 and the P-value is less than 0.0001. This indicates a highly significant deviation from the expected 9:3:3:1 ratio.

Question 2

To quantify linkage, Morgan crossed homozygous black, normal-winged females with homozygous gray, vestigial-winged males. He then crossed the F1 females with black, vestigial-winged males. (You should note that this is not the same test cross as the one shown in the experiment in Figure 12.15A. In that case, the original parents were BBVgVg and bbvgvg.) The results of this test cross are shown in the table (Experiment 2). Are these genes linked? If they are linked, what is the map distance between the genes? Explain why these data are so different from the data shown in Figure 12.15A.

The genes are linked, because there more parental type phenotypes and fewer recombinant (non-parental) phenotypes in the test cross progeny than would be expected for unlinked genes.

The map distance is:

[(578 + 307) / (578 + 307 + 1413 + 1117)]  x 100 =  26 map units

These data are different from the data shown in Figure 12.17, because the parental types are different: black normal and grey vestigial in this cross, and grey normal and black vestigial in Figure 12.17.

Question 3

In a third experiment, Morgan crossed two genetic strains of flies that were homozygous for the body color and wing genes. The F1 flies were all gray and normal-winged, and these were interbred. The results are shown in the table (Experiment 3). What were the genotypes and phenotypes of the original parents that produced the F1?

Number of progeny showing each phenotype
Experiment Gray,
normal		 Black,
normal		 Gray,
vestigial		 Black,
vestigial
1 2,316 1,146    737     0
2    578 1,413 1,117 307
3    246         9      65   18

BBVgVg and bbvgvg

A similar work with the data exercise may be assigned in LaunchPad.

Morgan knew about chromosomes and meiosis. To explain the data, he proposed that the two Drosophila loci are on the same chromosome—that is, that they might be linked. Suppose that the B and Vg loci are indeed located on the same chromosome. Why didn’t all of Morgan’s F1 flies have the parental phenotypes—that is, why didn’t his cross result in gray flies with normal wings and black flies with vestigial wings, in a 1:1 ratio? If linkage were absolute—that is, if chromosomes always remained intact and unchanged—we would expect to see just those two types of progeny. However, this does not always happen.