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Self Assessment Answer #
2
by Dr Jamie Love |
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Self Assessment Answer #
2
by Dr Jamie Love |
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I hope you remembered that a dihybrid cross should give you a 9 : 3 : 3 : 1 ratio with the 9 representing the "doubly dominant" genotypes (A-B-), the 3's representing those that are homozygous recessive at only one locus (either A-bb or aaB-) and the 1 representing the category of homozygous recessives at both loci (aabb).
The F2s had these phenotypes and in these numbers
Red and tall = 30
White and tall = 65
Red and short = 83
White and short = 206
The best way to match them up is to understand that they are arranged
1 : 3 : 3 : 9
It might help to write in better genotypes than the As and Bs I've been using.
Obviously white dominates red and short dominates tall.
(Just because tall dominated dwarf in the peas doesn't mean it will do so in roses!) Here's that data again with their genotypes -as near as we can tell.
Red and tall (wwss) = 30
White and tall (W-ss) = 65
Red and short (wwS-) = 83
White and short (W-S-) = 206
You don't have to work out these genotypes to do the chi-square but I thought this would be a good time to remind you how we figure them out and organize a chi-square for four categories.
It might also help to remind you of the chi-square equation.
2 = 
[(O - E)2/E] with "O" = "observed" and "E" = "expected".
The "" means to sum all the stuff inside the brackets. Previously, that has meant to sum only two values, representing the two categories but here we are working with four categories so you should understand that we will be adding up the final values for ALL four of them to get 2.
You are now educated well enough to be able to work the chi-square more efficiently and the equation might help.
We are given the observed number and must first calculate the expected. There is a total of 384 plants (30 + 65 + 83 + 206 = 384) but what is the ratio we would expect if the 1 : 3 : 3 : 9 ratio holds true here?
I think it's best to start with the doubly homozygous recessive group, red and tall (wwss), because that is the "1" in the ratio. They should represent 1/16 of the total population (1 + 3 + 3 + 9 = 16). [NOT 1/9! That is a common mistake.]
To determine the expected number of red, tall plants simply divide 384 by 16 to get 24. (382/16 = 24) That means we expected to get 24 tall, red plants instead of 30. The two in the middle (the ones that must be homozygous recessive for one locus but not the other) will represent 3/16 of the total population (or simply 3 times as many as the doubly homozygous recessive) so there should be 72 white, talls and 72 red, shorts instead of the 65 and 83 observed. Finally, there should be 216 white, short plants (W-S) in the F2s instead of the 206 observed.
Rather than walk you through each step of the chi-square I have summarized the math in the table below. In an exam, you might want to construct such a table because it allows you to work more effectively and quickly.
E |
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(wwss) |
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(W-ss) |
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(wwS-) |
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(W-S-) |
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Now we add the four values together (that's what "" tells us) to get a 2 = 4.325
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That is a very large number but let's see what the table tells us about it.
If you used one degree of freedom, as we have done with a monohybrid cross, you would have rejected this result because the calculated chi-square here (
There are four categories here, not two, so there are three degrees of freedom in this experiment . That means that the correct value to look at is 7.81. Our calculated |
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So we have no reason to reject this data as not representing a 9 : 3 : 3 : 1 ratio. It makes sense - statistical sense.
Chi-square rules! 
That was a difficult chi-square but it was also an important one because you have to practice these difficult problems in order to understand them. Maybe you missed that one and feel discouraged. Don't let it get you down. Instead, go back and do it again, now that you've seen how it is done, and try to understand where you went wrong and why.
Return to the Question.
