1. Expected Values and Observed Values Usually Differ
If I were to give you a coin and tell you to flip it 50 times, you’d expect the results to be about 50 heads and 50 tails. That’s a reasonable expectation, based on the fact that the coin has two sides, and your toss is a pretty good randomizer (assuming it’s an honest coin and that you make a vigorous toss: click here to read about how that isn’t always the case). But there’s a good chance that you won’t get 50/50. In four trials, you might observe:
 48 heads and 52 tails.
 45 heads and 55 tails
 57 heads and 43 tails
 60 heads and 40 tails
If the discrepancy is small (as in 48:52), it probably won’t bother you. You’ll say to yourself something like “This doesn’t bother me. It’s just random variation.” But there’s also a level at which the discrepancy becomes so great that it does bother you. You might start to wonder if something — another variable — is at play.
The difference between expected and observed results is a basic problem in the sciences. To solve this problem, scientists use a statistical method called the χ^{2} test. “χ” is the Greek letter “chi,” which (despite the “ch” at the start) is pronounced”kai” (and rhymes with “pie”). The test is pronounced “kai squared.”
Let’s see how it works.
2. Understanding the Null Hypothesis
Let’s start by getting some data from a genetic cross. We’ll use this to clarify some ideas related to the χ^{2} test, and then we’ll learn how to use the test.
Here’s the scenario: a group of students is doing a breeding experiment with the fruit fly, Drosophila melanogaster. The students breed a whiteeyed male (genotype X^{w}Y) with a redeyed heterozygous female (genotype X^{W}X^{w}). Note that these are sexlinked genes alleles: click here for a review.
[qdeck]
[h] Whiteeyed male crossed with heterozygous redeyed female
[q]1. Set up a Punnett square for a cross between a whiteeyed male (genotype X^{w}Y) and a redeyed heterozygous female, genotype X^{W}X^{w}
2. List the expected outcome for each phenotype.
[a]1. Here’s the Punnett square
X^{w}  Y  
X^{W}  X^{W}X^{w}  X^{W}Y 
X^{w}  X^{w}X^{w}  X^{w}Y 
2. Here are the expected results:
1 redeyed female: 1 whiteeyed female: 1 redeyed male: 1 whiteeyed male.
[/qdeck]
The students’ experiment produced a total of 476 offspring. Based on the Punnett square, we’d expect four phenotypes, each equally represented. That means we expect 119 individuals of each type (divide 476 by 4 to get 119). The table below shows the observed results, and the expected results immediately below
RedEyed Female  WhiteEyed Female  Redeyed male  Whiteeyed male  
Observed  109  113  137  117 
Expected  119  119  119  119 
There’s a difference between the observed and expected results. We can respond in two ways.
 We can say, “We’re okay with this difference. The difference is insignificant, random, and not important. Our expectation (hypothesis) was correct.” Saying this means that we accept what statisticians call the NULL HYPOTHESIS. The null hypothesis means that there’s no statistically significant difference between observed and expected results.
 We can say “We’re not okay with this difference.” In that case, we do not accept the null hypothesis and have to conclude that
 Our expectation was flawed, or
 Our experimental methods were flawed.
What are the criteria by which we decide when to accept the null hypothesis, and when not to? That’s exactly what the χ^{2} test is for.
3. How to do a χ^{2} test in 7 steps
The formula for the χ^{2} test is
Here’s what each part of the equation means:
 χ^{2} is “chisquared.” It’s what we’re solving for.
 Σ means “the sum of.”
 O means “observed.”
 E represents “expected.”
So, in plain English, the formula is
 χ^{2 }equals the sum of ((observed – expected) squared), divided by the expected))
I’ve tried to use parentheses to show the order of operations. But if it’s not clear, it will be after we try it out.
To avoid confusion, I always do χ^{2 } by setting up a table and then plugging in the numbers.
There are seven steps.
 STEP 1: Set up your table
 STEP 2: Enter the observed and expected values.
 STEP 3: In each column, subtract expected from observed: (O – E).
 STEP 4: In each column, square the value of observed – expected: (OE)^{2}
 STEP 5: In each column, divide (observed expected)^{2} by the expected: (OE)^{2}/E.
 STEP 6: Add the values of (OE)^{2}/E.
 STEP 7: Look up the value in a critical values table (discussed below) and decide whether to accept your null hypothesis.
Let’s apply these steps to the problem above.
STEP 1: Set up your table. The number of rows is always the same. The number of columns varies based on the number of categories. In this case, there were four phenotypes, so we need four columns.
RedEyed Female  WhiteEyed Female  Redeyed male  Whiteeyed male  
1. Observed (O)  
2. Expected (E)  
3. Observed – Expected (O – E)  
4. (OE)^{2}  
5. (OE)^{2}/E 
STEP 2: Enter the observed and expected values. Observed values are what you measured, tallied, etc. For the expected values, if you’re expecting equal representation, then just take the total and divide by the number of categories (as we’ll do here). If the expected ratios are not equal then you’d have to do a little more math to figure out your expected values. We’ll tackle that in the sample problems below.
RedEyed Female  WhiteEyed Female  Redeyed male  Whiteeyed male  
1. Observed (O)  109  113  137  117 
2. Expected (E)  119  119  119  119 
3. Observed – Expected (O – E)  
4. (OE)^{2}  
5. (OE)^{2}/E 
STEP 3: Subtract expected from observed: (O – E). This is the value in row 1 – the value in row 2 in each column. Your answer can be a positive or a negative value.
RedEyed Female  WhiteEyed Female  Redeyed male  Whiteeyed male  
1. Observed (O)  109  113  137  117 
2. Expected (E)  119  119  119  119 
3. Observed – Expected (O – E)  10  6  18  2 
4. (OE)^{2}  
5. (OE)^{2}/E 
STEP 4: In each column, square the value of observed – expected: (OE)^{2}. This is the value in row 3 of each column, multiplied by itself.
RedEyed Female  WhiteEyed Female  Redeyed male  Whiteeyed male  
1. Observed (O)  109  113  137  117 
2. Expected (E)  119  119  119  119 
3. Observed – Expected (O – E)  10  6  18  2 
4. (OE)^{2}  100  36  324  4 
5. (OE)^{2}/E 
STEP 5: In each column, divide (observed expected)^{2} by the expected: (OE)^{2}/E. This is the value in row 4 divided by the value in row 2.
RedEyed Female  WhiteEyed Female  Redeyed male  Whiteeyed male  
1. Observed (O)  109  113  137  117 
2. Expected (E)  119  119  119  119 
3. Observed – Expected (O – E)  10  6  18  2 
4. (OE)^{2}  100  36  324  4 
5. (OE)^{2}/E  100/119=
0.84 
36/119 =
0.30 
324/119 =
2.72 
4/119 =
0.03 
STEP 6: Add the values of (OE)^{2}/E. These are all the values in row 5, added together.
0.84 + 0.13 + 2.72 + 0.03 = 3.89
So, for this data set χ^{2 } = 3.89
STEP 7: Look up the value in a critical values table and decide whether or not to accept your null hypothesis.
The table below is a critical values table. Here’s how to use it.

 Determine the number of categories in your data set. In a genetics problem, that’s the number of phenotypes.
 Take the number of categories, and subtract 1. That’s your Degrees of Freedom (df)
 Find the table cell where probability (pvalue) at the 0.05 level meets your degrees of freedom. Why 0.05? As someone who’s not a statistician (or even a scientist), I can only say that p = 0.05 is a widely agreedupon standard for what constitutes statistical significance. As a student in an AP Bio course, that’s all you need to know. However, if you want to dig deeper into these issues, consult this page from the Statistics Teacher website.
 Determine if your χ^{2 } value is less than the p = 0.05 value at the degrees of freedom for your problem.
 If it is less, then you ACCEPT the null hypothesis. That means that you interpret the difference between your observed and expected values as not being significant.
 If it isn’t less then you cannot accept the null hypothesis. You have to rethink your expectation or your experimental design.
Let’s apply these steps to the problem above.
 We have four categories (male with red eyes, male with white eyes, female with red eyes, and female with white eyes).
 Four categories – 1 = 3. That’s our degrees of freedom.
 The pvalue for 0.05 meets 3 degrees of freedom at 7.82.
 Our χ^{2 } was 3.89. Because 3.89 is less than 7.82, we accept our null hypothesis: the difference between observed and expected values was not statistically significant.
That’s it. Let’s make sure we’ve got the basic ideas, and then we’ll do some practice problems.
4. χ^{2:} Checking Understanding
[qwiz qrecord_id=”sciencemusicvideosMeister1961Chi Square Checking Understanding (v2.0)”] [h]
χ^{2: }Key Terms and Concepts
[q] χ^{2 }is a statistical technique for evaluating the importance of the difference between [hangman] values (the ones that you measure) and [hangman] values.
[c]IG9ic2VydmVk[Qq]
[f]IENvcnJlY3Qh[Qq]
[c]IGV4cGVjdGVk[Qq]
[f]IEV4Y2VsbGVudCE=[Qq]
[q] The claim that the difference between observed and expected values is not statistically significant is known as the [hangman] [hangman].
[c]IG51bGw=[Qq]
[f]IEdvb2Qh[Qq]
[c]IGh5cG90aGVzaXM=[Qq]
[f]IEV4Y2VsbGVudCE=[Qq]
[q] An educator is studying the relationship between ontime attendance and grades. She divides the students who are the subject of a study into the following groups: early, on time, and tardy. After she gathers data and does her χ^{2} test, how many degrees of freedom will there be?
[textentry single_char=”true”]
[c]ID I=[Qq]
[f]IFllcy4gVGhlcmUgYXJlIHRocmVlIGNhdGVnb3JpZXMgYW5kIHR3byBkZWdyZWVzIG9mIGZyZWVkb20u[Qq]
[c]ICo=[Qq]
[f]IE5vLiBIZXJlJiM4MjE3O3MgYSBoaW50LiBUaGUgZGVncmVlcyBvZiBmcmVlZG9tIGlzIHRoZSBudW1iZXIgb2YgY2F0ZWdvcmllcyAtMS4gSG93IG1hbnkgY2F0ZWdvcmllcyBhcmUgdGhlcmU/[Qq]
[q] In peas, the allele for purple flowers (P) is dominant over the allele for white flowers (p). If you carry out a monohybrid cross, and then statistically analyze the results using a χ^{2} test, how many degrees of freedom will there be?
[textentry single_char=”true”]
[c]ID E=[Qq]
[f]IFllcy4gVGhlcmUgYXJlIG9ubHkgdHdvIHBoZW5vdHlwZSBjYXRlZ29yaWVzLCBwdXJwbGUgYW5kIHdoaXRlLCBzbyB0aGUgZGVncmVlcyBvZiBmcmVlZG9tID0gMS4=[Qq]
[c]ICo=[Qq]
[f]IE5vLiBIZXJlJiM4MjE3O3MgdGhlIG1vbm9oeWJyaWQgY3Jvc3M6
Cg==Cg==Cg==Cg==[Qq]  P  p 
P  PP  Pp 
p  Pp  pp 
P is dominant. How many phenotypic categories will there be? Take that number, subtract 1, and you’ll have the degrees of freedom.
[q multiple_choice=”true”] A student is evaluating the phenotypic results of a dihybrid cross: AaBb x AaBb. If they want to evaluate their phenotypic results at the 0.05 probability level, what critical value do they use?
[c]IDMuODQ=[Qq]
[f]IE5vLiBBIGRpaHlicmlkIGNyb3NzIHdpbGwgcmVzdWx0IGluIGZvdXIgcGhlbm90eXBpYyBjYXRlZ29yaWVzOg==
Cg== Cg==
 RG9taW5hbnQgZm9yIGJvdGggdHJhaXRzOw== Cg==
 [Qq]Dominant for the first trait, recessive for the second;
 Recessive for the first trait, dominant for the second;
 Recessive for both traits.
The number of degrees of freedom is the number of categories – 1. Find where the pvalue of 0.05 intersects with that many degrees of freedom, and you’ll have the answer.
[c]IDUuOTk=[Qq]
[f]IE5vLiBBIGRpaHlicmlkIGNyb3NzIHdpbGwgcmVzdWx0IGluIGZvdXIgcGhlbm90eXBpYyBjYXRlZ29yaWVzOg==
Cg== Cg==
 RG9taW5hbnQgZm9yIGJvdGggdHJhaXRzOw== Cg==
 [Qq]Dominant for the first trait, recessive for the second;
 Recessive for the first trait, dominant for the second;
 Recessive for both traits.
The number of degrees of freedom is the number of categories – 1. Find where the pvalue of 0.05 intersects with that many degrees of freedom, and you’ll have the answer.
[c]IDcu ODI=[Qq]
[f]IEV4Y2VsbGVudC4gQSBkaWh5YnJpZCBjcm9zcyB3aWxsIHJlc3VsdCBpbiBmb3VyIHBoZW5vdHlwaWMgY2F0ZWdvcmllczo=
Cg== Cg==
 RG9taW5hbnQgZm9yIGJvdGggdHJhaXRzOw== Cg==
 [Qq]Dominant for the first trait, recessive for the second;
 Recessive for the first trait, dominant for the second;
 Recessive for both traits.
The number of degrees of freedom is the number of categories – 1. In this case, it’s 4 categories, so 3 degrees of freedom. The pvalue of 0.05 intersects with 3 degrees of freedom at 7.82.
[c]IDkuNDk=[Qq]
[f]IE5vLiBBIGRpaHlicmlkIGNyb3NzIHdpbGwgcmVzdWx0IGluIGZvdXIgcGhlbm90eXBpYyBjYXRlZ29yaWVzOg==
Cg== Cg==
 RG9taW5hbnQgZm9yIGJvdGggdHJhaXRzOw== Cg==
 [Qq]Dominant for the first trait, recessive for the second;
 Recessive for the first trait, dominant for the second;
 Recessive for both traits.
The number of degrees of freedom is the number of categories – 1. Find where the pvalue of 0.05 intersects with that many degrees of freedom, and you’ll have the answer.
[q multiple_choice=”true”] A student is evaluating the phenotypic results of a test cross: AABb x aabb. If they want to evaluate their data at the 0.05 probability level, what critical value do they use?
[c]IDMu ODQ=[Qq]
[f]IE5pY2Ugam9iLiBUaGUgdGVzdCBjcm9zcyBiZXR3ZWVuIEFBQmIgYW5kIGFhYmIgcmVzdWx0cyBpbiB0d28gZ2Vub3R5cGVzIGFuZCB0d28gcGhlbm90eXBlczo=
Cg== Cg==
 QWFCYjogRG9taW5hbnQgZm9yIGJvdGggdHJhaXRzOw== Cg==
 [Qq]Aabb: Dominant for the first trait, recessive for the second trait.
The number of degrees of freedom is the number of categories – 1. That’s one degree of freedom. The pvalue of 0.05 intersects with one degree of freedom at 3.84.
[c]IDUuOTk=[Qq]
[f]IE5vLiBUaGUgdGVzdCBjcm9zcyBiZXR3ZWVuIEFBQmIgYW5kIGFhYmIgcmVzdWx0cyBpbiB0d28gZ2Vub3R5cGVzIGFuZCB0d28gcGhlbm90eXBlczo=
Cg== Cg==
 QWFCYjogRG9taW5hbnQgZm9yIGJvdGggdHJhaXRzOw== Cg==
 [Qq]Aabb: Dominant for the first trait, recessive for the second trait.
The number of degrees of freedom is the number of categories – 1. Find where the pvalue of 0.05 intersects with that many degrees of freedom, and you’ll have the answer.
[c]IDcuODI=[Qq]
[f]IE5vLlRoZSB0ZXN0IGNyb3NzIGJldHdlZW4gQUFCYiBhbmQgYWFiYiByZXN1bHRzIGluIHR3byBnZW5vdHlwZXMgYW5kIHR3byBwaGVub3R5cGVzOg==
Cg== Cg==
 QWFCYjogRG9taW5hbnQgZm9yIGJvdGggdHJhaXRzOw== Cg==
 [Qq]Aabb: Dominant for the first trait, recessive for the second trait.
The number of degrees of freedom is the number of categories – 1. Find where the pvalue of 0.05 intersects with that many degrees of freedom, and you’ll have the answer.
[c]IDkuNDk=[Qq]
[f]IE5vLlRoZSB0ZXN0IGNyb3NzIGJldHdlZW4gQUFCYiBhbmQgYWFiYiByZXN1bHRzIGluIHR3byBnZW5vdHlwZXMgYW5kIHR3byBwaGVub3R5cGVzOg==
Cg== Cg==
 QWFCYjogRG9taW5hbnQgZm9yIGJvdGggdHJhaXRzOw== Cg==
 [Qq]Aabb: Dominant for the first trait, recessive for the second trait.
The number of degrees of freedom is the number of categories – 1. Find where the pvalue of 0.05 intersects with that many degrees of freedom, and you’ll have the answer.
[q multiple_choice=”true”] In incomplete dominance, heterozygotes have an intermediate phenotype. A scientist is performing a statistical analysis of the outcome of a monohybrid cross for a trait that involves incomplete dominance. If they want to evaluate their data at the 0.05 probability level, what critical value do they use?
[c]IDMuODQ=[Qq]
[f]IE5vLiBJbiBpbmNvbXBsZXRlIGRvbWluYW5jZSwgYSBtb25vaHlicmlkIGNyb3NzIChzdWNoIGFzIHRoYXQgYmV0d2VlbiBBYSBhbmQgQWEpIHdpbGwgcmVzdWx0IGluIHRocmVlIHBoZW5vdHlwZXM6
Cg== Cg==
 QUE6IERvbWluYW50IHBoZW5vdHlwZQ== Cg==
 [Qq]Aa: Intermediate phenotype
 aa: Recessive phenotype
The number of degrees of freedom is the number of categories – 1. Find where the pvalue of 0.05 intersects with that many degrees of freedom, and you’ll have the answer.
[c]IDUu OTk=[Qq]
[f]IE5pY2Ugam9iISBJbiBpbmNvbXBsZXRlIGRvbWluYW5jZSwgYSBtb25vaHlicmlkIGNyb3NzIChzdWNoIGFzIHRoYXQgYmV0d2VlbiBBYSBhbmQgQWEpIHdpbGwgcmVzdWx0IGluIHRocmVlIHBoZW5vdHlwZXM6
Cg== Cg==
 QUE6IERvbWluYW50IHBoZW5vdHlwZQ== Cg==
 [Qq]Aa: Intermediate phenotype
 aa: Recessive phenotype
The number of degrees of freedom is the number of categories – 1. In this case, 3 phenotypes means 2 degrees of freedom. The pvalue of 0.05 intersects with that many degrees of freedom at 5.99.
[c]IDcuODI=[Qq]
[f]IE5vLiBJbiBpbmNvbXBsZXRlIGRvbWluYW5jZSwgYSBtb25vaHlicmlkIGNyb3NzIChzdWNoIGFzIHRoYXQgYmV0d2VlbiBBYSBhbmQgQWEpIHdpbGwgcmVzdWx0IGluIHRocmVlIHBoZW5vdHlwZXM6
Cg== Cg==
 QUE6IERvbWluYW50IHBoZW5vdHlwZQ== Cg==
 [Qq]Aa: Intermediate phenotype
 aa: Recessive phenotype
The number of degrees of freedom is the number of categories – 1. Find where the pvalue of 0.05 intersects with that many degrees of freedom, and you’ll have the answer.
[c]IDkuNDk=[Qq]
[f]IE5vLiBJbiBpbmNvbXBsZXRlIGRvbWluYW5jZSwgYSBtb25vaHlicmlkIGNyb3NzIChzdWNoIGFzIHRoYXQgYmV0d2VlbiBBYSBhbmQgQWEpIHdpbGwgcmVzdWx0IGluIHRocmVlIHBoZW5vdHlwZXM6
Cg== Cg==
 QUE6IERvbWluYW50IHBoZW5vdHlwZQ== Cg==
 [Qq]Aa: Intermediate phenotype
 aa: Recessive phenotype
The number of degrees of freedom is the number of categories – 1. Find where the pvalue of 0.05 intersects with that many degrees of freedom, and you’ll have the answer.
[/qwiz]
5. χ^{2} Practice Problems
At this point, you should be ready for some practice problems. Set up your data in tables, systematically follow the seven steps I laid out above, and you’ll be fine.
[qdeck scroll = “true” style=” minheight: 600px !important; width: 675px !important;” card_back=”none” qrecord_id=”sciencemusicvideosMeister1961Chi Square Practice Problems (v2.0)”]
[h] χ^{2} Practice Problems
[q] In corn, purple seeds are dominant over yellow seeds. Smooth seeds are dominant to wrinkled. In an experimental dihybrid cross, the following results were obtained.
Phenotype  Purple Smooth  Purple Wrinkled  Yellow Smooth  Yellow Wrinkled 
Observed  190  60  79  18 
Run a χ^{2} test to determine if the difference between observed and expected is significant or not. Your answer 1) will list a χ^{2} value, 2) interpret the results in a way that accepts or fails to accept the null hypothesis, with a justification of your interpretation.
[a] Because this is a dihybrid cross, you’re expecting a 9:3:3:1 ratio. Of the 347 offspring, that means that you expect 9/16 to have both dominant traits. So, multiply 347 by 9/16, and you have 195.19 that you’re expecting with that phenotype. Determine your other expected numbers by multiplying the total by 3/16 and 1/16.
Purple, smooth  Purple wrinkled  yellow, smooth  yellow, wrinkled  total  
observed  190  60  79  18  347 
expected  195.2  65.1  65.1  21.7  
oe  5.2  5.1  13.9  3.7  
(oe)(oe)  26.9  25.6  194.3  13.6  
((oe)(oe))/e  0.14  0.39  2.99  0.63  4.14 
The χ^{2} value is 4.14. There are 4 categories, meaning 3 degrees of freedom, so the critical value at the 0.05 level is 7.82. Because 4.14 is below 7.82, we can accept the null hypothesis. The difference is not statistically significant.
[q] In Drosophila, the whiteeyed trait is a recessive, sexlinked mutation. In a breeding experiment, a whiteeyed male is crossed with a heterozygous, redeyed female. The following results were obtained:
phenotype  male, white eyes  male, normal eyes  female, white eyes  female, normal eyes 
observed  86  109  115  133 
Run a χ^{2} test to determine if the difference between observed and expected is significant or not. Your answer 1) will list a χ^{2} value, 2) interpret the results in a way that accepts or fails to accept the null hypothesis, with a justification of your interpretation.
[a] If you make this Punnett square…
X^{w}  Y  
X^{W}  X^{W}X^{w}  X^{W}Y 
X^{w}  X^{w}X^{w}  X^{w}Y 
Running a χ^{2} test on the data above gives you the following:
phenotype  male, white eyes  male, normal eyes  female, white eyes  female, normal eyes  total 
observed  86  109  115  133  443 
expected  110.8  110.8  110.8  110.8  
oe  24.8  1.8  4.3  22.3  
(oe)(oe)  612.6  3.1  18.1  495.1  
(oe)(oe)/e  5.5  0.0  0.2  4.5  10.2 
χ^{2} = 10.2. There are 4 categories, meaning 3 degrees of freedom, so the critical value at the 0.05 level is 7.82. Because 10.2 is above 7.82, we cannot accept the null hypothesis. The difference is statistically significant, and (since we know that our hypothesis is correct), we need to look for additional variables that are causing the discrepancy.
[q] In Drosophila, apterous is a mutation that causes flies to develop extremely reduced or absent wings. The mutation is autosomal and recessive. Two flies, both heterozygous normal, are bred together, with the following results.
Phenotype  Normal wings  apterous 
Observed  643  189 
 Using the symbols ap^{+} and vg, create a Punnett square and make a hypothesis about expected results.
 Run a χ^{2} test to determine if the difference between observed and expected is significant or not. Your answer 1) will list a χ^{2} value, 2) interpret the results in a way that accepts or fails to accept the null hypothesis, with a justification of your interpretation.
[a] 1. If you make this Punnett square…
ap+  ap  
ap+  ap+ ap+  ap+ ap 
ap  ap+ ap  ap ap 
…you’d expect 3/4 normal winged, and 1/4 apterous
That leads to the following χ^{2} test on the data above gives you the following:
male, normal wings  female, apterous  total  
observed  643  189  832 
expected  624.0  208.0  
oe  19.0  19.0  
(oe)(oe)  361.0  361.0  
(oe)(oe)/e  0.6  1.7  2.3 
The χ^{2} value is 2.3. There are 2 categories, meaning 1 degree of freedom, so the critical value at the 0.05 level is 3.84. Because 2.3 is below 3.84, we can accept the null hypothesis. The difference is not statistically significant.
[q] In Drosophila, sepia (se) is a mutation that produces brown eyes. This allele is autosomal and recessive to the wildtype allele (se^{+}), which produces red eyes. Apterous (ap) is an autosomal recessive mutation resulting in reduced or absent wings. The wildtype allele is ap^{+}).
Students in a lab perform a dihybrid cross. In addition to looking at the phenotypes for sepia and apterous, they also categorize their flies by sex. Here are their results.
male, normal wing, normal eye  female, normal wing, normal eye  male, apterous, normal eye  female, apterous, normal eye  male, normal wing, sepia eyes  female, normal wing, sepia eyes  male, apterous, sepia  female, apterous, sepia  
observed  196  358  57  90  85  117  16  21 
 Make a hypothesis about expected results.
 Run a χ^{2} test to determine if the difference between observed and expected is significant or not. Your answer 1) will list a χ^{2} value, 2) interpret the results in a way that accepts or fails to accept the null hypothesis, with a justification of your interpretation.
[a] 1. This is a dihybrid cross, so you’re expecting a 9:3:3:1 ratio. But adding sex, you’re expecting 9/32 male wild type, 9/32 female wild type, 3/32 male normal wings, sepia eyes, 3/32 female normal wings sepia eyes, etc.
2. Your χ^{2} test is as follows:
male, normal wing, normal eye  female, normal wing, normal eye  male, apterous, normal eye  female, apterous, normal eye  male, normal wing, sepia eyes  female, normal wing, sepia eyes  male, apterous, sepia  female, apterous, sepia  total  
observed  196  358  57  90  85  117  16  21  940 
expected  264.4  264.4  88.1  88.1  88.1  88.1  29.4  29.4  
oe  68.4  93.6  31.1  1.9  3.1  28.9  13.4  8.4  
(oe)(oe)  4675.1  8765.6  968.8  3.5  9.8  833.8  178.9  70.1  
(oe)(oe)/e  17.7  33.2  11.0  0.0  0.1  9.5  6.1  2.4  79.9 
χ^{2} = 79.9. There are 8 categories, meaning 7 degrees of freedom, so the critical value at the 0.05 level is 15.51. Because 79.9 is above 15.51, we cannot accept the null hypothesis. The difference is statistically significant, and (since we know that our hypothesis is correct), we need to look for additional variables that are causing the discrepancy.
[/qdeck]
What now?
Continue to Topics 5.3 – 5.5, Part 7: Mitochondrial Inheritance, Incomplete Dominance, and GenotypeEnvironment Interaction, the next tutorial in AP Bio Unit 5.