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During his studies on genes in Drosophila that were sex linked, T. H. Morgan found $F_2$ population phenotypic ratios deviated from expected 9:3:3:1. Explain the conclusion he arrived at.

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Last updated date: 18th Jun 2024
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Answer
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Hint: Morgan picked the organic product fly, Drosophila melanogaster, for his hereditary examinations. What organic product flies may need charm (contingent upon your intuition regarding insects), they compensate for much in reality: they're modest, simple, and quick to develop. As it turned out, Morgan affirmed Mendelian laws of legacy and the theory that qualities are situated on chromosomes.

Complete answer:
Conclusion of Morgan's investigations are:
-Genes were situated on the X-chromosome.
-When the two qualities in a dihybrid cross were concentrated on a similar chromosome, the extent of parental quality blends were a lot higher than the non-parental type.
Morgan expressed this relationship as 'linkage' to portray the physical relationship of qualities on a chromosome.
Connected genes happen on the same chromosome while unlinked genes happen in various chromosomes. Connected genes do not show autonomous arrangements. They do not give a 9:3:3: dihybrid phenotypic proportion and 1:1:1:1 as a twofold test cross proportion.
-Morgan and his group likewise found that in any event, when genes were assembled on a similar chromosome, a few of them were firmly connected (low recombination) and others were approximately connected (high recombination).

Note:
T. H. Morgan saw that when the two qualities in a dihybrid cross are situated on a similar chromosome, the extent of parental quality blends in the offspring was a lot higher than the non-parental or recombination of genes. Morgan and his associates found that when qualities were gathered on a similar chromosome, a few genes are firmly connected and show less recombination. At the point when the genes are inexactly connected, they show higher recombination.