Biology

Using a Marble Model

Materials

Box containing 100 marbles of two colors

Introduction

Working in pairs, you will test Hardy‑Weinberg equilibrium by simulating a population using colored marbles. The box of marbles represents the gene pool for the population. Each marble should be regarded as a single gamete, the two colors representing different alleles of a single gene. Each box should contain 100 marbles of the two colors in the proportions specified by the instructor. Record in the space provided below the color of the marbles and the initial frequencies for your gene pool.

A = ​(color) allelic frequency

a = (color) allelic frequency

(p + q)2 = p2 + 2(pq) + q2 = 1

(0.5 + 0.5)2 = 0.52 + 2(0.5 x 0.5) + 0.52 = 1

1 = 0.25 + 0.50 + 0.25 = 1 (all genes in the population)

1. There are 100 alleles in your box, how many diploid individuals are represented in this population?

2. What would be the color combination of the marbles needed to produce a homozygous dominant individual?

3. What would be the color combination of the marbles needed to produce a homozygous recessive individual?

4. What would be the color combination of the marbles needed to produce a heterozygous individual?

Hypothesis

State the Hardy‑Weinberg theorem in the space provided. This will be your hypothesis (it is sort of a null hypothesis…assuming no selection, etc.)

Predictions

Predict the genotypic frequencies of the population in future generations (if/then). Deductive thinking.

PART I Procedure-

1. Without looking, randomly remove two marbles from the box. These two marbles represent one diploid individual in the next generation. In the table to the right, record a tally of the diploid genotype (AA, Aa, or aa) of the individual formed from these two gametes.

2. Return the marbles to the box and shake the box to reinstate the gene pool. By replacing the marbles each time, the size of the gene pool remains constant and the probability of selecting any allele should remain equal to its frequency. This procedure is called sampling with replacement.

3. Repeat steps 1 and 2 (select two marbles, record the genotype of the new individual, and return the marbles to the box) until you have recorded the genotypes for 50 individuals who will form the next generation of the population.

AA individuals	Aa individuals	aa individuals

PART I Results-

1. Before calculating the results of your experiment , determine the expected frequencies of genotypes and alleles for the population. To do this, use the original allelic frequencies for the population provided by the instructor. (Recall that the frequency of A = p, and the frequency of a = q.) Calculate the expected genotypic frequencies using the Hardy‑Weinberg equation p2 + 2pq + q2 = 1 . The number of individuals expected for each genotype can be calculated by multiplying 50 (total population size) by the expected frequencies. Record these results in Table 12.1.

Table 12.1 Expected Genotypic and Allelic Frequencies for the Next Generation Produced by the Marble Model

Parent Populations	EXPECTED New Populations
Allelic Frequency	Genotypic Number (# individuals) and Frequency (proportion)	Allelic Frequency
A	a	AA # = Freq.=	Aa # = Freq.=	aa # = Freq.=	A	a

2. Next, using the results of your experiment , calculate the observed frequencies in the new population created as you removed marbles from the box. Record the number of diploid individuals for each genotype in Table 12.2, and calculate the frequencies for the three genotypes (AA, Aa, aa). Add the numbers of each allele, and calculate the allelic frequencies for A and a. These values are the observed frequencies in the new population. Genotypic frequencies and allelic frequencies should each equal 1.

Table 12.2 Observed Genotypic and Allelic Frequencies for the Next Generation Produced by the Marble Model.

Parent Populations	OBSERVED New Populations
Allelic Frequency	Genotypic Number (# of individuals) and Frequency (proportion)	Allelic Frequency
A	a	AA # = Freq.=	Aa # = Freq.=	aa # = Freq.=	A	a

3. To compare your observed results with those expected, you can use a chi-square test of the genotype frequencies . Table 12.3 will assist in the calculation of the chi-square test. Fill out this table, but then use excel to perform a chi-squared test and calculate an actual p-value.

Table 12.3 Chi-Square of Results from the Marble Model

	# of AA individuals	# of Aa individuals	# of aa individuals
Observed value (o)
Expected value (e)
Deviation (o – e) = d
d2
d2/e
Calculated Chi-square value (X2) = Σd2/e

Degrees of freedom (# of possible genotypes – 1) = __________, P-value from excel chi-square test = ____________

Are your observed genotypic frequencies the same, or significantly different, than what is expected according to H-W Equilibrium? Why?

PART II Procedures-

1. Follow the same procedures as in PART I, except this time keep your eyes open as you select the marbles. Decide on one of the two colors to have a “selective advantage”, then make your selections favoring choosing that color. (You don’t need to choose them every time, but don’t choose the marbles randomly.)

2. Record your tallies for the new population below, then calculate your “observed frequencies” and record them in Table 12.4, below.

AA individuals	Aa individuals	Aa individuals

3. Disregard the fact that you did not select the marbles randomly, and use your original “expected frequencies” from Table 12.1 (according to H-W Equilibrium) to run a chi-square test on the genotype frequencies, using excel .

PART II- Results

Table 12.4 Observed Genotypic and Allelic Frequencies for the Next Generation Produced by the (non-random) Marble Model.

Parent Populations	OBSERVED New Populations
Allelic Frequency	Genotypic Number (# of individuals) and Frequency (proportion)	Allelic Frequency
A	a	AA # = Freq.=	Aa # = Freq.=	aa # = Freq.=	A	a

Chi-square test p-value:

Are your observed genotypic frequencies the same, or significantly different, than what is expected according to H-W Equilibrium? Why?

Post-lab 5 Questions

(detach and turn in next week along with your abstract)

1. In your PART I newly produced generation, what proportion of your population was…

a. Homozygous dominant?

b. Homozygous recessive?

c. Heterozygous?

2. Do your results in PART I match your predictions for a population at Hardy-Weinberg equilibrium?

3. If you continued the PART I simulation for 25 generations…

a. What would you expect to happen to the frequencies of each allele?

b. Would that population be evolving? Explain your response.

4. Consider each of the conditions for the Hardy‑Weinberg model. Does your model meet each of those conditions?

Consider what you did in PART II:

5. How did the simulation in PART II differ from your methods in PART I?

6. Which condition necessary for H-W Equilibrium was violated?

7. What did your chi-square test allow you to identify about the genotype frequencies in the new population in PART II?

8. Is the population in PART II evolving? Explain your response.

9. How could this method be useful when studying population genetics of real organisms?

10. What might be some of the difficulties you would encounter while using this in a real population genetics study?

For example, if the “A” allele has a frequency of 0.75 in a gene pool, then p = 0.75, and the frequency of “a” can be calculated as q = 0.25.

If “A” has a frequency of p = 0.75 the probability of getting an “AA” genotype is:

p2 = 0.752 = 0.5625

Likewise for “a”, with q = 0.25, probability of “aa” is:

q2 = 0.252 = 0.0625

Prob. of “Aa” or “aA” is:

2pq = 2(0.75 x 0.25) = 0.375

NOTE:

p2 + 2pq + q2 = 1

0.5625+0.375+0.0625 =1