10.2 Dihybrid crosses and gene linkage 10.2.1 Calculate and predict the genotypic and phenotypic ration of offspring of dihybrid crosses involving unlinked autosomal genes. •
Example: In salamanders, spots are dominant to solid bodies, and red is dominant to blue coloring. Predict the phenotypic ratio of the offspring if one parent is heterozygous for spots and homozygous recessive for color, and the other parent is heterozygous. o Start by writing the genotypes of each parent: Ssrr x SsRr o Draw a 16 square grid and place the possible allele combinations of one parent on the top, and the combinations of the other parent on the side. As you can see from the table below, the parent on the top only really has two possible allele combinations, and the parent on the side has four possibilities. Sr sr Sr sr SR Sr sR sr o Then fill in the grid with the corresponding allele combinations. Sr sr Sr sr SR SSRr SsRr SSRr SsRr Sr SSrr Ssrr SSrr Ssrr sR SsRr ssRr SsRr ssRr sr Ssrr ssrr Ssrr ssrr o Now write down the possible phenotypic combinations as in the table below. o Count the number of individuals that have each of the combinations. o As you can see, the two columns on the right are the same as the columns on the left, so I am only going to count the two columns on the left. This will be the case whenever either of the parents are homozygous for either of the traits. Phenotype combination Spots and Red (dominant for both) Spots and blue (dominant for one recessive for other) solid and Red (dominant for one recessive for other) solid and blue (recessive for both) Sr sr Sr sr SR SSRr SsRr SSRr SsRr Sr SSrr Ssrr SSrr Ssrr sR SsRr ssRr SsRr ssRr sr Ssrr ssrr Ssrr ssrr Number of individuals in the cross that have this combination. 3 3 1 1 o The ratio is 3:3:1:1 Spots & Red: Spots and blue: solid and Red: solid and blue 10.2.2 Distinguish between autosomes and sex chromosomes. •
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Sex chromosomes are the pair that determines gender. A female has the pair XX, and male XY. The X chromosome is much larger and carries a number of necessary genes, the Y chromosome carries the SRY gene that will stimulate testosterone production and repress the female specific traits. The other 22 pairs are autosomes. 10.2.3 Explain how crossing over between non-­‐sister chromatids of a homologous pair in prophase 1 can result in an exchange of alleles. •
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Each autosome carries thousands of genes. Homologous pairs carry the same genes, but might have different alleles for the same gene. Lets use salamanders for the example again. As it turns out, the gene for spots is found on the same chromosome as the gene for eye color. Black eyes are dominant to green eyes. We are going to look at one salamander that is heterozygous for both traits. The dominant alleles are carried on the same chromosome, and the recessive alleles are carried on the same chromosome. •
During prophase 1 of meiosis 1, the homologous pairs cross over. This can happen multiple times between the non-­‐sister chromatids, in our example we are showing just one. •
After crossing over there are new combinations of alleles, each new chromatid will go into a different gamete. 10.2.4 Define linkage group. •
Genes carried on a particular chromosome are called a linkage group. The genes for eye color and spots in the example above are part of a linkage group. 10.2.5 Explain an example of a cross between two linked genes. •
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A classic example is with fruit flies (Drosophila). In Drosophila, body color and wing shape are carried on the same gene (are part of a linkage group). The gene for body color comes in two alleles, the wild type is dominant with a gray body (b+ is the symbol used for this allele). The mutant color is recessive with a black body (b). The gene for wing shape comes in two alleles, the wild type dominant of normal wings (vg+) and the mutant recessive is vestigial wings (vg). Consider an individual that is heterozygous. For this individual the wild type genes are found on the same chromosome, and the mutant genes are found on its homologous pair. After meiosis is complete that might not be the case due to crossing over (see the picture example above and see if you can explain it using the fly example). 10.2.6 Identify which of the offspring are recombinants in a dihybrid cross involving linked genes. •
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In the following example you can see that the P generation starts with two flies, one homozygous for the wild type traits, the other homozygous for the mutant traits. All of their offspring (F1 generation) will be heterozygous for both traits. That F1 hybrid will then be crossed with a double mutant. By crossing it with a double mutant you can see the frequency of crossing over between the homologous chromosomes in the F1 hybrid. Anytime one of their offspring has a different combination of phenotypes than one of the two parents, they are considered a recombinant. The fewer recombinants, the more closely linked the genes are, which means the more closely they are located on the chromosome. In fact, the percentage of recombinants is translated to map units. 25% recombinants would mean 25 map units on a chromosome map. •
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Close to 50% recombinants would indicate that the genes are far enough apart that they actually behave like genes on separate chromosomes. In the same example below you can follow how the genes behave in the cross between the F1 hybrid and the mutant. As you can see the percentage of recombinants is 17%, which means the chromosomes are located 17 map units apart. •
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For IB you will need to be able to determine the possible recombinants for a test cross. Test crosses always cross a fly with another fly that is recessive for the trait or traits being investigated. So in this example the test cross is between the individual below and a double mutant (double recessive) fly. The double recessive fly’s genes are not indicated. b+ vg+ the F1 hybrid, shows the dominant phenotypes for each trait. The testcross is with a fly with recessive phenotypes. b vg b+ vg b vg+ Theses possible recombinants show the one dominant & phenotype and one recessive phenotype. b vg b vg