We took a sample of five rows, and counted the different types of kernels. We choose five rows because, though it may be slightly less accurate, it was easier to do with the amount of time given. Our data is listed below.
Now, we counted the number of each specific combination of traits in the five rows of the corn. Our results are listed below.
Luckily, we obtained the expected 9:3:3:1 ratio, which means that the genes are located on the same chromosome and do assort independently. This 9:3:3:1 ratio is the expected ratio between the two heterozygous parents. We created a punnett square of the parents to ensure that the expected ratio is correct.
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Dark Purple = Purple and Smooth (9)
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Now, time for the Chi Square part. We calculated the individual chi square values for each row and added them all together to determine the overall chi square value. We once again used a table:
Then, we determined whether the chi square value is a good fit with our data. We were told that the degrees of freedom (df) is the number of possible phenotypes - 1. We then circled the row on a chart that was closest to our chi square value.
The closest value is between .58 and 1.42 under the "Good Fit Between Ear & Data" section. "Good fit" and "poor fit" are referring to how your observed ratio is to the expected ratio of 9:3:3:1 that would result from the PsSs parents. Our data was surprisingly close to the expected value, however two reasons the data would have a poor chi square value are if you get unlucky and choose a dud set of rows that don't exhibit the correct ratio, or you simply counted wrong.
Chi Square Problem Set:
1. Problem: A large ear of corn has a total of 433 grains, including 271 Purple & starchy, 73 Purple & sweet, 63
Yellow & starchy, and 26 Yellow & sweet.
Your Tentative Hypothesis: This ear of corn was produced by a dihybrid cross
(PpSs x PpSs) involving two pairs of heterozygous genes resulting in a theoretical (expected) ratio of 9:3:3:1
Objective: Test your hypothesis using chi square and probability values.
The chi square value was in the poor fit section, meaning that the that the genes are not located on the same chromosome or do not assort independently. Therefore, my hypothesis is wrong.
2. Problem: In a certain reptile, eyes can be either black or yellow. Two black eyed lizards are crossed, and the result
is 72 black eyed lizards, and 28 yellow-eyed lizards.
Your Tentative Hypothesis: The black eyed parents were Bb x Bb
Objective: Test your hypothesis using chi square analysis. In this set, because only two values (traits) are examined,
the degrees of freedom (df) is 1. SHOW ALL WORK!
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Expected:75% Black
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My hypothesis is correct because the chi square value lands in the "good fit" section, meaning the parents are both heterozygous black (Bb).
3. Problem: A sample of mice (all from the same parents) shows
58 Black hair, black eyes 16 Black hair, red eyes
19 White hair, black eyes 7 White hair, red eyes
Your tentative hypothesis: (what are the parents?)
Objective: Use a chi square analysis to support your hypothesis
Total = 100
Black hair = dominant (H)
Black eyes = dominant (E)
Hypothesis: parents are heterozygous black hair, black eyes. HhEe and HhEe
Expected ratio would be 9:3:3:1
The chi square value lands in the "Good Fit," meaning that my hypothesis is true and the parents are heterozygous black hair, black eyes (HhEe).
Conclusion
This lab activity made my understanding of chi square much more clear. It seems tough at first, but as long as you remain organized and show all your data in a neat chart, it's surprisingly easy. Aside from chi square, this lab has taught me to be an adept user of excel and to have the patience to count hundreds of corn kernels. I hope Mr Wong has uses similar labs to explain hard concepts because I believe that hands-on work is how I learn best.
