5. In the tomato the mutant genes o (oblate flattened fruit), p (peach = hairy fruit), and s (compound inflorescence = many flowers in a cluster) were found to be linked on chromosome 2. A testcross with a heterozygote was done, yielding the following progeny:. || || || || +++ = 73 ++s = 348 +p+ = +ps = o++ = 110 o+s = op+ ops = va 382 = 609 96 2 306 63 Determine a) the sequence and arrangement of the genes in the heterozygous parent; and b) the distance between the three genes.

Gene mapping is a basic procedure in genetics used to find and determine the general location of a chromosome. It gives information about linear order of genes, the distances between them, and their linkage relationships. Gene mapping should be possible through different techniques, including linkage mapping, which tracks the inheritance pattern of genes in families, and physical mapping, which pinpoints genes' actual position on a chromosome.This data is significant for grasping the hereditary basis of characteristics, infections, and different biological cycles. Gene planning has been instrumental in propelling fields like medication, agriculture, and evolutionary biology by empowering scientists to investigate the hereditary landscapes of the living world.Understanding the information provided in the question text:To determine the sequence and arrangement of genes on the chromosome, as well as the distance between the three genes (o, p, and s), we can perform a gene mapping analysis in light of the test cross results.We should initially recognize the gene plan in the heterozygous parent, which ought to give us data about which quality is in the center. In view of the test cross progeny:+++ = 73 (wild type for all genes)++s = 348 (homozygous for s, so it's at the end)+p+ = 2 (homozygous for p, so it's at one end)+ps = 96 (heterozygous for p, so it's between p and s)From the above results, we can infer that the gene arrangement is:o - p - sNow, to calculate the distance between the genes, we need to consider the recombinant offspring, as they are the result of crcrossing-overvents. The following recombinant progeny is observed:op+ = 306ops = 63The recombination frequency is the ratio of recombinant progeny to the total number of offspring.Recombination frequency (RF) = (Number of recombinant offspring) / (Total number of offspring)For o and p genes,the recombinants are op+ and ops,and the total number of offspring is 306 + 63 = 369.RF(op+) = 306 / 369 ≈ 0.828RF(ops) = 63 / 369 ≈ 0.171Now, you need to understand that the recombination frequency is directly proportional to the physical distance between genes on a chromosome.In a centiMorgan (cM) scale, this is equivalent to percentages. So, the distance between o and p is approximately 82.8 cM, and between p and s, it's approximately 17.1 cM.The genes o, p, and s are arranged in the following sequence on chromosome 2:o - p (82.8 cM) - s (17.1 cM)Hence the final answer will be,the sequence of arrangement of genes in homozygous parent - o - p - sAnd, the distance between o and p is approximately 82.8 cM, and between p and s, it's approximately 17.1 cM.

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