Question about biology - gene mapping

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SUMMARY

The discussion focuses on calculating recombination frequency (RF) to determine the linkage between genes in genetic mapping. The formula RF = (#Recombinants)/(Total offspring) x 100 is emphasized as a critical tool for this analysis. Participants highlight that a recombination frequency of 50% or higher indicates unlinked genes, while lower frequencies suggest linkage. The example provided illustrates how to analyze progeny phenotypes to assess gene linkage effectively.

PREREQUISITES
  • Understanding of genetic terminology, including "recombination frequency" and "linkage."
  • Familiarity with basic genetics concepts, such as genotype and phenotype.
  • Ability to perform calculations involving ratios and percentages.
  • Knowledge of Mendelian inheritance patterns.
NEXT STEPS
  • Study the principles of genetic linkage and mapping in more detail.
  • Learn how to perform chi-square tests for genetic data analysis.
  • Explore software tools for genetic mapping, such as MapMaker or JoinMap.
  • Investigate the implications of gene linkage on inheritance patterns in various organisms.
USEFUL FOR

Students in genetics courses, educators teaching genetic mapping, and researchers involved in genetic analysis will benefit from this discussion.

andydan1056
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Homework Statement


Screen Shot 2015-08-02 at 8.17.22 PM.png


Homework Equations


RF = (#Recombinants)/(Total offspring) x 100

The Attempt at a Solution


How do I know that b is unlinked? I know that unlinked genes have a recombination frequency of 50% or higher, but how do I calculate RF for the genes here?
 
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andydan1056 said:

Homework Statement


View attachment 86734

Homework Equations


RF = (#Recombinants)/(Total offspring) x 100

The Attempt at a Solution


How do I know that b is unlinked? I know that unlinked genes have a recombination frequency of 50% or higher, but how do I calculate RF for the genes here?

Practiced geneticists and maybe you by the end of these exercises or others can just look at these numbers and say ah yes this and that are linked in this order, but we have to reason it out bit by bit, remembering what linkage means.

To start you off, I would say e.g.the female is heterozygous so has genotype Aa so should have 50% A and 50% a progeny.
But don't rely on it, count them, in fact I find 500/1000 have phenotype + and same number are a.
Phenotype + is genotype Aa (AA should not be possible), phenotype a genotype aa.
Is a linked to b?
If it is unlinked, 50% of the + that have A should have B and be + for the b phenotype, a and b inherited independently.
If it is closely linked then either 100% or 0% of the a should be + for b.
I counted up just the + for a and found 254/500 of the +a were +b.
So loci a and b appear essentially unlinked.
You should also look at those that were a. Presumed genotypically aa.
So do all the combinations like that, That should give you the linkages or not, at least for the autosomals
 

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