Stickleback Evolution Virtual Lab | HHMI's BioInteractive

Stickleback Evolution Lab Progress

Summary | Quizzes | Data | Analysis

Experiment 1: Chi-Square Analysis

Your Pelvic Structure Scores Totals

Lake population

Pelvic spines present

Pelvic spines absent

Bear Paw Lake:

Frog Lake:

Morvoro Lake:

Your Chi-Square Calculations

Lake Population

x²

Bear Paw Lake:

Frog Lake:

Morvoro Lake:

1. Regarding the null hypothesis, which combination of results is correct?

Answer to Chi-Square Calculations

Lake Population

x²

Bear Paw Lake:

0.01

Frog Lake:

0.01

Morvoro Lake:

0.8

0.5

Bear Paw Lake = fail to reject; Frog Lake = fail to reject; Morvoro Lake = reject
Bear Paw Lake = fail to reject; Frog Lake = reject; Morvoro Lake = fail to reject
Bear Paw Lake = reject; Frog Lake = reject; Morvoro Lake = fail to reject
Bear Paw Lake = fail to reject; Frog Lake = reject; Morvoro Lake = reject

Why?

The p value for the data for Bear Paw Lake and Frog Lake is less than 0.05, which means that the null hypothesis (i.e., your expectation of 50:50) can be rejected. The p value for the data for Morvoro Lake is greater than 0.05; thus, the null hypothesis cannot be rejected.

2. What is one plausible explanation for the results in question 1?

Our results show that it is likely that both the population in Bear Paw Lake and the one in Frog Lake differ from the null hypothesis.
Based on our results, we can claim without a doubt that all populations having stickleback without pelvic spines occur in lakes with the same ecological conditions as Bear Paw Lake.
Based on our results, we know that stickleback from Morvoro Lake are not statistically different from those in Bear Paw Lake.
Our results show that the stickleback populations in Bear Paw and Frog lakes are unlikely to differ from the null hypothesis, and so any apparent differences are strictly due to sample error.

Why?

The null hypothesis is that we expect roughly equal frequencies of each pelvic phenotype, presumably because there is no selective pressure acting on any one form of the trait. This hypothesis can be rejected for Bear Paw Lake and Frog Lake, and there could be an ecological explanation for this finding. However, studying only two lake populations is not enough to generalize to other stickleback populations. In 1990, Dr. Bell and colleagues sampled more than 200 freshwater sites (mostly lakes but also streams and rivers) around Cook Inlet. They found that pelvic reduction occurs at substantial frequency (p ≥ 5%) in lakes that lack predatory fishes, and that this association was unlikely to be due to sample error. They concluded that pelvic reduction evolved separately in different lakes due to natural selection. (The results were published in Bell, M. A., et al. 1993. Evolution of pelvic reduction in threespine stickleback fish: a test of competing hypotheses. Evolution 47:906-914.)

3. In biological research, what does a probability (p-value) of 0.01 mean?

A p-value of 0.01 does not offer any meaningful value for interpreting the results.
A p-value of 0.01 means that If the null hypothesis is true, there is a 1% chance of getting a difference between the observed and expected values as big or bigger than the one we detected.
A p-value of 0.01 means that if the null hypothesis is true, there is a 99% chance of getting a difference between the observed and expected values as big or bigger than the one we detected.
A p-value of 0.01 means that there is a 1% chance that this difference can be explained by the alternative hypothesis.

Why?

In this context, the p-value is the probability that the observed result is really different from the null hypothesis. In a chi-square test, only the null hypothesis can be rejected. Failure to reject the null hypothesis is support for the alternative hypothesis; the test itself does not directly test the alternative hypothesis. All we can say is that there is only a 1% chance that we would incorrectly reject the null hypothesis, if it is true.

4. What does it mean when a null hypothesis is rejected?

There is no statistically significant difference between the observed and expected data.
There is no statistically significant difference between the observed and expected data. Therefore, the difference is due exclusively to chance.
There is a difference between the observed and expected data and it is unlikely that the difference happened by chance.
Rejection of the null hypothesis tells us why pelvic reduction has occurred because the alternative hypothesis has been proven true.

Why?

An alternative hypothesis is never proven true with any statistical test like the chi-square test. This statistical test tells you whether the null hypothesis can or cannot be rejected.

Experiment 2: Rate of Change Analysis

The graph you constructed in part 3 of this experiment shows that, over time, the frequency of fish with a complete pelvis decreases as the frequency of fish with a reduced pelvis increases. In this portion of the lab, you calculated the rate that best describes how quickly this change occurred in the population.

Your Rate-of-Change Calculations
Time	Rate of change per thousand years
First 3,000 years (From layer 1 to layer 2)
Next 3,000 years (From layer 2 to layer 3)
Next 3,000 years (From layer 3 to layer 4)
Next 3,000 years (From layer 4 to layer 5)
Next 3,000 years (From layer 5 to layer 6)

1. When did the greatest decrease in the frequency of pelvic spines occur?

Answer to Rate-of-Change Calculations
Time	Rate of change per thousand years
First 3,000 years (From layer 1 to layer 2)	-3%
Next 3,000 years (From layer 2 to layer 3)	-25%
Next 3,000 years (From layer 3 to layer 4)	-3%
Next 3,000 years (From layer 4 to layer 5)	-1.7%
Next 3,000 years (From layer 5 to layer 6)	0%

Between layers 4 and 5
Between layers 1 and 2
Between layers 2 and 3
Between layers 5 and 6

Why?
The complete-pelvis phenotype decreased by about 25% every 1,000 years from layer 2 to layer 3. This was the most rapid decrease.

2. Which of the following statements best describes the trend over time:

Answer to Rate-of-Change Calculations
Time	Rate of change per thousand years
First 3,000 years (From layer 1 to layer 2)	-3%
Next 3,000 years (From layer 2 to layer 3)	-25%
Next 3,000 years (From layer 3 to layer 4)	-3%
Next 3,000 years (From layer 4 to layer 5)	-1.7%
Next 3,000 years (From layer 5 to layer 6)	0%

The percentage of fish with a complete pelvis in this population decreased at a constant rate between layers over the 15,000-year period examined.
Pelvic reduction evolved rapidly in the first 9,000 years and then slowly leveled off toward the end of the 15,000-year period, at which point no fish with a complete pelvis appear.
There is an increase in the percentage of fish with a complete pelvis, and this increase is most rapid in the first 9,000 years.
You would need many more fish to draw any conclusions about trends over time.

Why?

The best way to visualize the rate of change over time is to graph the data, which might look something like the graph below. Note that the time interval between each layer is the same. The graph shows that the complete-pelvis phenotype decreased during the first 9,000 years. The greatest decrease occurred between 3,000 and 6,000 years (layers 2 and 3) after colonization of the lake by a population of fish with a complete pelvis.

3. Which of the following statements best describes why a researcher would want to calculate the rate of change of a particular trait in fossil fish?

Answer to Rate-of-Change Calculations
Time	Rate of change per thousand years
First 3,000 years (From layer 1 to layer 2)	-3%
Next 3,000 years (From layer 2 to layer 3)	-25%
Next 3,000 years (From layer 3 to layer 4)	-3%
Next 3,000 years (From layer 4 to layer 5)	-1.7%
Next 3,000 years (From layer 5 to layer 6)	0%

Calculating the rate of change of a trait in fossil fish usually helps researchers identify the ecological factors (i.e., predation or competition) acting as agents of selection on the specified trait.
Rate of trait change calculations are only useful for practicing math skills.
By calculating the rate of change of a trait, you can infer the original size of the fish population that lived in that lake.
Calculating rates of change in fossil fish can be used to infer some aspects of the evolution of traits in living fish, which are not accessible for study because of the long time scale of most evolutionary processes.

Experiment 3: Chi-Square Analysis

Your Score Totals

Bear Paw Lake

Coyote Lake

Left bias:

Right bias:

Your Chi-Square Calculations

Lake Population

x²

Bear Paw Lake:

Coyote Lake:

1. For this experiment, the null hypothesis was that there should be no preferred bias in pelvic asymmetry. In other words, in the two populations you studied you would expect a roughly equal number of fish with a left bias as with a right bias. Based on the chi-square calculations, which result is correct?

Answer to Chi-Square Calculations

Lake Population

x²

Bear Paw Lake:

12.8

0.01

Coyote Lake:

0.01

We cannot reject the null hypothesis for either Bear Paw Lake or Coyote Lake.
We cannot reject the null hypothesis for Bear Paw Lake but we can for Coyote Lake.
We can reject the null hypothesis for both Bear Paw Lake and Coyote Lake.
We can reject the null hypothesis for Bear Paw Lake but not Coyote Lake.

Why?

The probability is less than 0.05 for both lakes. This means that the null hypothesis (i.e., your expectation of 50:50) can be rejected for Bear Paw Lake and Coyote Lake.

2. What can we infer from rejecting the null hypothesis in this particular experiment?

There could be a biological explanation for why stickleback in both Bear Paw Lake and Coyote Lake have pelvic vestiges that are larger on the left side.
Based on our results, we can conclude that all stickleback populations have left-biased asymmetry.
Based on our results, we know that some environmental factor must cause the pelvic vestiges to be larger on the left side.
Any difference between the results obtained in Bear Paw Lake and Coyote Lake is due to sample error.

Why?

The null hypothesis is that we expect roughly equal frequencies of left- and right-biased pelvic asymmetry because the asymmetry should occur by chance. This hypothesis can be rejected for both Bear Paw Lake and Coyote Lake. As a result there could be a biological explanation for why stickleback in these two populations have a pelvic vestige that is larger on the left than the right side. We don’t know whether there is an environmental factor that acts on this phenotype, but based on this analysis we know that there could be some explanation and that the results are probably not due to pure chance.