120

A flask of nutrient broth, buffered to maintain pH, is inoculated with a strain of E. coli. The flask is placed in a constant temperature environment where it is aerated by shaking.

A. Predict the effect of a change in energy availability over time.

B. Represent the change graphically in terms of the number of cells as a function of time.

C. In your graph as time progresses there is a change in the growth rate of the population. Add annotation to your graph to describe the time interval during which the growth rate is increasing linearly in proportion to the number of cells. Add annotation to your graph to describe another time interval during which the growth rate is decreasing in proportion to the square of the number of cells. Add a third annotation to describe an interval of time where the rate of growth is zero.

D. Select and justify two measurements of the E. coli population that could be made at two different points in time during growth that would be sufficient to answer questions about the population size at any time.

E. Describe the population of E. coli if the environment was continuously supplement by additional nutrient broth.

121

The following problem extends the Hardy-Weinberg model of population dynamics that was covered in Chapter 19. It applies mathematics that would be appropriate after a second course in Algebra. While the concept applied in this problem are within the scope of the Exam the mathematical representations are not and the item is provided to allow students who are able another look at the concepts.

The Hardy-Weinberg model of population dynamics is an algebraic representation of the relationships among genotype frequencies, F, and the probability of the dominant allele A, p, and the recessive allele a, q. The Hardy-Weinberg model of population dynamics is based on several assumptions. One of these assumptions is “random mating.” If all genes in a population are equally able to reproduce, this means that all genes are equally fit and equally fertile. Consequently, the population never evolves.

Populations do evolve and the Hardy-Weinberg model can be modified slightly to allow evolution to occur. Suppose that there is an initial population at generation zero and the probability of the dominant allele at that time is p0. Later, at population k the probability is different. But if the frequencies of the three different combinations of alleles is known then the probabilities pk and qk can be calculated at generation k

(1)

http://www.w3.org/1998/Math/MathML" display="block">pk=Fk(AA)+&#xBD;Fk(Aa)qk=Fk(aa)+&#xBD;Fk(Aa)pk=Fk(AA)+½Fk(Aa) qk=Fk(aa)+½Fk(Aa)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">pk=Fk(AA)+½Fk(Aa)qk=Fk(aa)+½Fk(Aa)

And since p and q are probabilities for a case where only two alleles exist, p+q=1. Then also (p+q)2=1, leading the Hardy-Weinberg equation

http://www.w3.org/1998/Math/MathML" display="block">p2k+2pkqk+q2k=1p2k+2pkqk+q2k=1" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">p2k+2pkqk+q2k=1

The gene distribution never changes and pk=pk-1.

The equations of the Hardy-Weinberg model were modified (Haldane, 1924) to create a model in which evolution occurs:

(2)

http://www.w3.org/1998/Math/MathML" display="block">Fk(AA)=p2kwAA/WFk(Aa)=2pkqkwAa/WFk=Fk(AA) = p2kwAA/W Fk(Aa) = 2pkqkwAa/W Fk =" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">Fk(AA)=p2kwAA/WFk(Aa)=2pkqkwAa/WFk=

http://www.w3.org/1998/Math/MathML" display="block">q2kwaa/WW=p2wAA+2pqwAa/q2waa q2kwaa/W W = p2wAA+2pqwAa/q2waa" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">q2kwaa/WW=p2wAA+2pqwAa/q2waa

Haldane divides by the factor W=Fk(AA)+Fk(Aa)+Fk(aa) so that the probabilities that are still calculated with equation (1) to continue to satisfy the condition for p and q to represent probabilities: (p+q)2=1.

A. Justify Haldane’s model in terms of what the factors wAA, wAa, and waa mean.

B. Suppose that wAA = wAa = 1, but that waa = 0.8. Predict what will happen to the population over time.

Fitness is determined by the environment. Moree (The American Naturalist, 86, 1952) measured the relative fitness in Drosophila melanogaster of a recessive allele that imparts black eye color as population density increases. A varying number of flies with an equal number of males and females were placed in a pint jar and progeny counted. In each experiment the population was initially heterozygous.

Number of females x Number of maleswaa
1 x 1 0
10 x 10 0.06
50 x 50 0.11
150 x 150 0.46
Table36.3

C. Apply Haldane’s approach to calculate the probability p in the first generation after mating 150 female and 150 male flies that are heterozygous using wAA = wAa = 1.

Rendel (Evolution, 5, 1951) conducted an investigation of the dependence of fecundity (fertility) on light in ebony-eyed D. melanogaster. A summary of some of the data that he reported is shown in the table below:

Fraction females inseminated
Phenotype of male Light condition Dark condition
Ebony 0.215 0.607
Wild type 0.494 0.466
Table36.4

D. Pose two scientific questions concerning the behavioral response indicated by the data that can be tested experimentally.

E. Is there a question you can add here to wrap up this set with this LO from the list? In this case “light” is the single environmental factor, and they two phenotypes are ebony and wild type that result from different genotypes within the population of flies.