How Do You Calculate Genotype and Phenotype Ratios in a Trihybrid Cross?

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SUMMARY

The discussion focuses on calculating genotype and phenotype ratios in a trihybrid cross, specifically crossing TTBbRr with TtbbRr. Participants emphasize the use of Punnett squares for visualizing outcomes and the importance of understanding the probability of each genotype and phenotype. The example provided illustrates how to derive ratios from a simpler Aa x Aa cross, demonstrating that outcomes can be calculated by counting occurrences of each genotype and dividing by the total number of outcomes. The consensus is that while Punnett squares are effective, using probability calculations can simplify the process for more complex crosses.

PREREQUISITES
  • Understanding of Mendelian genetics
  • Familiarity with Punnett squares
  • Basic knowledge of genotype and phenotype definitions
  • Ability to perform probability calculations
NEXT STEPS
  • Learn how to construct Punnett squares for trihybrid crosses
  • Study the principles of probability in genetics
  • Explore advanced genetic concepts such as epistasis and linkage
  • Investigate software tools for genetic simulations
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Students studying genetics, educators teaching Mendelian inheritance, and researchers interested in genetic probability calculations.

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Homework Statement



How would you figure out the genotypes and phneotypes if you Cross TTBbRr with TtbbRr?
Trihybrid cross


Homework Equations




Cross TTBbRr with TtbbRr




The Attempt at a Solution



I did the crossing, but i don't know how to do the genotypes nor the phenotype ratio
 
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\frac{Number \left \left of \left \left outcomes \left \left with \left \left a \left \left particular \left \left phenotype/genotype}{Total \left \left number \left \left of \left \left outcomes}
For example, if you performed Aa x Aa, then there are four possible outcomes, AA, Aa, aA and aa. These are all equally likely, and two are the same. So, for AA, it appears once as an outcome of four possible outcomes, thus you get 1/4. For, Aa (aA is the same), is appears twice out of four possible outcomes, thus 2/4, or 1/2. For aa, it appears once out of four possible outcomes, thus 1/4.
I think that's what you were looking for. For crosses where a large number of genes is being considered, punnett squares and writing out the possibilities can be tiresome; and just using probabilities is easier (although possibly more prone to error).
 

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