Counting Corn

In the 1600’s, people had no video games, so instead they had various activities such as husking bees, where if you get a red ear of corn, something good happens to you. But why only red years of corn? Mainly because there were not that much. Because red was the recessive gene. But since nobody knew that because Mendel was born in 1822 so he didn’t exist yet to publish his genetics theory. Now forward this 400 years to where we are now, and instead of having yellow dominant ears, we are counting purple dominant yellow recessive ears of corn for biology class(for points and our grade, not for kisses). We want to see the ears of corn and determine if the ears of corn are a good fit for Mendel’s theory or not. So we first counted the phenotypes of the kernels on five rows of a Monohybrid Cross, which should be a 3:1 ration for purple to yellow and the kernels should be all smooth.. Down below I have our diagram and picture:

Next we took a Test cross. Our expected was equal number for all types. Here is our diagram and picture:

Then we put our data into a chi square:

It has a 20-50% chance of the difference caused by chance, so therefore, it’s a good fit. So then our hypothesis of it being a 1:1:1:1 ratio is a good hypothesis and should not be tossed away. A good fit means how well the model(the corn) fits a set of observations. A poor fit is when you have to reconsider if there is a faulty approach and the hypothesis might be incorrect. A data might have poor chi square fit when Genes are linked or on the same chromosome, dramatically increasing the chance of a phenotype appearing. Another is when there is faulty data or just plain really really bad luck.

Now we have a problem set to finish.

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 total number of kernels is 840. So the expected would be:

Purple and Smooth 9/16 * 433 = 243.6

Purple and Wrinkled 3/16 * 433 = 81.2

Yellow and Smooth 3/16 * 433 = 81.2

Yellow and Wrinkled 1/16 * 433 = 27.1

so after calculating that out would be:

3.08 + .82 + 4.07 + .04 = 8.01. According to the chi square, this is a bad fit.

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!

If we draw a Punnet Square Bb and Bb, we should get a ratio of 3:1.

The total is 100, so we use that to calculate:

Black eyes should be 75 but is 72

Yellow eyes should be 25 but is 28

Calculating that out I get of chi value of .12+.36=.48 which is a good fit.

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?)

Since the result is about 9:3:3:1, this is probably a dihybrid cross, making its parents HhEe and HhEe. The total offspring is 100, so the expected is:

9/16 of 100 = 56.25

3/16 of 100 = 18.75

3/16 of 100 = 18.76

1/16 of 100 = 6.25

I calculated this out to get:

.05 + .40 + .003 + .09 = .54

On the chi square, this is a good fit.