In the tomato, Solanum esculentum, tall (D−)(D−) is dominant to dwarf (dd) plant height, smooth fruit (P−) is dominant to peach fruit (pp), and round fruit shape (O−) is dominant to oblate fruit shape (oo). These three genes are linked on chromosome 1 of tomato in the order dwarf–peach–oblate. There are 12 map units between dwarf and peach and 17 map units between peach and oblate. A trihybrid plant (DPO/dpo) is test-crossed to a plant that is homozygous recessive at the three loci (dpo/dpo). The accompanying table shows the progeny plants. Identify the mechanism responsible for the resulting data that do not agree with the established genetic map.

Question

In the tomato, Solanum esculentum, tall (D−)(D−) is dominant to dwarf (dd) plant height, smooth fruit (P−) is dominant to peach fruit (pp), and round fruit shape (O−) is dominant to oblate fruit shape (oo). These three genes are linked on chromosome 1 of tomato in the order dwarf–peach–oblate. There are 12 map units between dwarf and peach and 17 map units between peach and oblate. A trihybrid plant (DPO/dpo) is test-crossed to a plant that is homozygous recessive at the three loci (dpo/dpo). The accompanying table shows the progeny plants. Identify the mechanism responsible for the resulting data that do not agree with the established genetic map.

Table showing progeny phenotypes and their counts from a genetic cross in tomatoes.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem together identify the incorrect statement about the genetic implications of recombination in inversion heterozygote. The answer options are a, the probability of crossover within the inversion loop is linked to the size of the inversion loop. B. Inversion suppresses the production of recombinant chromosomes. C inversion, heterozygote only produce nonviable gametes. And option D fertility may be altered if an inversion, heterozygote carries a very large inversion. Now recall that an inversion heterozygote is an individual with one normal chromosome and one with an inversion chromosome. Now remember that in the inversion is when a segment of a chromosome is rotated 100 and 80 degrees on the screen. I am putting up a picture that illustrates both types of inversions and the products they produce after inversion loops. So let's discuss them. On the left hand side is a example of an perce paracentric inversion. The chromosomes are represented by lines, the normal chromosome. In this example is represented by a red brown line and has gene order ABC D. The inversion chromosome is depicted below it with gene order A B DC and it is represented by a blue line. Now remember in this type of inversion, the centromere is not involved. So our centromere is depicted by the black dots and are on the far and left of the chromosome lines before gene A. Now, after crossing over and an inversion loop, the products are depicted. The first product is a normal chromosome with a normal line or chromosomal length. And the JEAN order is ABC D. The second is a chromosome that is recombinant and contains a deletion of portions of genes C and D. Now, the gene order is still ABC D but the overall length and thus the overall length of the line depicted is shortened. The third product is also a recombinant chromosome and contains only gene A. So it has a large deletion and the fourth and final chromosome is the inversion chromosome with JEAN order A B DC. So of the products produced the recombinant chromosomes have deletions of genetic information and thus will produce nonviable gametes. Now, on the right hand side depicted is the paracentric inversion. The normal chromosome again is depicted by a red brown line and has an order gene A gene B, the centromere gene C and G gene D, the inversion chromosome is represented by a blue line and has order gene ac, the centromere and then gene B and D. So in this type of version, the centromere is involved and the inverted genes are genes C and B. The products after crossing over are shown. The first is a normal chromosome with the normal length and thus a normal line length and has order gene A B, the centromere and then gene C and D. The 2nd and 3rd chromosomes or products of crossing over contain duplications and deletions. They are the recombinant chromosomes. Now, the second chromosome, the order of genes for it is ABC A and the centro mirror is located in between genes B and C. The order of genes for the third chromosome is DB CD and the centromere is in between genes B and C. Now these two are the recombinant chromosomes and because they have duplication and deletion of genetic information, they will also produce nonviable gametes. And the final, the final product in a paracentric inversion is the inversion chromosome. So it has gene order ac, then the centromere and then gene B and D. So in both types of inversions of the products, there's four products for each one and in each type of inversion, half of the chromosomes will contain vi or will go on to produce viable gametes. So the normal and the inversion will produce viable gametes for both paracentric and perent inversions. While the recombinant chromosomes will produce nonviable Gams. Now, having reviewed all this information, let's identify the incorrect statement option. A that the probability of crossover within the inversion loop is linked to the size of the inversion loop is a true statement. The larger the inversion, the greater the probability of crossover occurring within the inversion loop. So option A is not the correct incorrect statement. Option B is also true. After crossing over occurs between homologous chromosomes, the recombinant gamings will be nonviable since they contain the deletions and or duplications depending on the type of inversion. And so inversions do suppress production of recombinant chromosomes. So option B is not the incorrect statement. Option D is also true. If there is a large inversion and it spans the majority of a chromosome, there's increased probability of crossing over occurring within the inversion. And again, this would result in two out of the four crossover products to go on to make nonviable gamines. That means half of the game meats are viable and half are not. So that greatly affects fertility. And so option D is not the incorrect statement. The incorrect statement is option C inversion, heterozygote only produce nonviable gams. All right, everyone. I hope you found this helpful and I'll see you soon for the next practice problem.

Verified video answer for a similar problem:

Key Concepts

Genetic Linkage

Map Units and Recombination Frequency

Test Cross