How about a general rule?
As you know, 2 genes in a dihybrid cross give 4 different phenotypes. How can you determine the likelihood of any given phenotype? Here is a general procedure:
1. Find all cells in the Punnett Table corresponding to that phenotype.
2. Find the likelihood of those cells by multiplying the probabilities associated with the mother's and father's contributions (i.e., the row and column probabilities) using the Law of AND.
3. Find the total probability of the phenotype by adding all the cell probabilities using the Law of OR.
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OK, so let's give it a try: click on each button below to find the probability of that phenotype, assuming a 10% recombination frequency.
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A 'regular' (non-recombinant) double-hybrid cross results in a 9:3:3:1 ratio -- or, if you had 100 babies, this is the same as a 56:19:19:6 ratio (approximately). (Math note: the 9:3:3:1 ratio assumes 16 babies, and you can convert this to a ratio involving 100 total babies by multiplying by 100/16).
The recombinant double-hybrid cross results in a 70:5:5:20 ratio. Quite a difference (that is, more wild-type kids and more double-variant kids, but fewer of the middle types).
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