Calculations in AP Biology
Calculations in AP Biology!
• Water Potential
• Rates (population growth, reaction rates, …)
• Hardy-Weinberg Calculations (allele frequencies, genotype frequencies, phenotype
frequencies)
• Chi Square Calculations (Statistical Interpretation of data – the same or not the same)
• Graphing Analysis, Standard Deviation, Standard Error of the Mean (SEM)
1. The ability to taste PTC is governed by a single autosomal gene with alleles T and t. The
trait to taste PTC is dominant. In a survey of 400 graduate students, 336 were found to be
tasters.
a. Determine the percentage of non-tasters in this population?
0.16 = 16%
b. Determine the frequency of the t allele in this population?
0.40 = 40%
c. Determine the frequency of the heterozygous genotype in this population.
2*0.60*0.40 = 0.48 = 48%
d. Imagine a hypothetical scenario in which the population is exposed to a new, deadly
poison that is very similar in chemical structure to PTC and can be detected only by tasters.
i. In the presence of this poison, is the population in Hardy-Weinberg equilibrium? If
not, which condition is violated.
In the presence of the poison, the population is not in Hardy-Weinberg
equilibrium. Because there would presumably be an advantage in
detecting/tasting the poison, tasters would have a selective advantage and
would live to reproduce at a greater rate than non-tasters.
ii. How do you expect the allele frequency will change over time?
The frequency of the T allele will increase, and the frequency of the t
allele will decrease.
iii. Does this represent evolution of the population?
Yes, this change in allele frequencies over time is evolution.
2. In an experiment to determine the concentration of solutes in carrot cytoplasm, carrot cores
were soaked in glucose solutions of various concentrations.
Concentration
% Change in
The carrot cores were massed prior to soaking, and were then
(M)
Mass
immersed in the glucose solution for 24 hours, removed from
0.00
+19.3
the solution, patted dry, and remassed. The percent change in
the mass of the carrot cores was then calculated. The data are
0.20
+5.5
given in the table at right.
0.40
-2.2
0.60
-10.6
i. Use the data to create an appropriate graph on the axes
0.80
-15.0
provided.
1.00
-17.8
%Change Mass
+20.0
+10.0
0.0
-10.0
-20.0
0.00
0.20
0.40
0.60
0.80
1.00
Concentration (M)
ii. Calculate the water potential of the 0.40 M glucose solution. Include appropriate units
with your response. (Assume that the temperature is 298 K and use R = 0.0831 L.bar/mol.K)
Ψs = -iCRT = -1*0.40 mol/L * 0.08314 bar/mol.K * 298 K = -9.9 bar
iii. Interpolate from your graph to determine the concentration of a glucose solution that is
isotonic to the carrot cell.
0.36 M (see graph)
iv. Calculate the solute potential of the carrot cell. Again, include appropriate units.
Ψs = -iCRT = -1*0.36 mol/L * 0.08314 bar/mol.K * 298 K = -8.9 bar
v. The carrot core that was immersed in pure water (0.00 M glucose) was soaked for several
more days. After each day, the core was removed, patted dry, and weighed. It was found that
the percent mass increase stabilized at +22.0% and that the carrot core was very turgid.
Calculate the pressure exerted by the cell wall on the cytoplasm under these conditions.
Ψtot = Ψp + Ψs
Ψtot = 0
Ψs = -8.9 bar
Ψs = +8.9 bar
3. In corn, the trait for tall plants (T) is dominant to the trait for short plants (t), and the trait for
yellow kernels (Y) is dominant to the trait for white kernels (y). In an experiment to study the
genetics of these traits, a student crossed true-breeding, homozygous dominant tall, yellowkerneled plants with short, white kerneled plants to obtain the F1 generation of plants.
Plants from the F1 generation were crossed with true-breeding short, white-kerneled plants to
obtain the F2 generation. The table below represents the predicted phenotypes of the F2
generation, along with the data obtained.
F2 Generation Phenotypes
Tall
Yellow Kernel
Tall
White Kernel
Short
Yellow Kernel
Short
White Kernel
a. Complete the table at right, showing
the genotypes of the P and F1 generation:
Number Expected
512
512
512
512
P
Number Observed
675
344
373
656
TTYY
F1
X
ttyy
TtYy
b. Write a Punnett square showing the expected distribution of phenotypes in the F2
generation.
ty
TY
Ty
tY
ty
TtYy
Ttyy
ttYy
ttyy
The expected phenotypes are ¼ tall, yellow-kernel, ¼ tall, white kernel, ¼
short, yellow-kernel, and ¼ short, white-kernel.
c. Explain the Expected Number of each phenotype. What assumption is made regarding the
distribution of alleles for plant height and kernel color?
The expected number of each phenotype assumes that the alleles for
height and kernel-color sort independently. If the genes are linked (on the
same chromosome, then the traits will not be sorted independently and there
will be some correlation between the traits.
CHI-SQUARE TABLE
d. Is the difference in expected outcomes and
observed outcomes due to chance? Support your
conclusion with a calculation.
χ =∑
2
Degrees of Freedom
p
1
2
3
4
0.05
3.84
5.99
7.82
9.49
0.01
6.64
9.32
11.34
13.28
(o - e )2
e
(675 - 512)2 + (344 - 512)2 + (373 - 512 )2 + (656 - 512)2
=
512
512
512
512
= 51.89 + 55.12 + 37.74 + 40.5
= 185.25 > 7.82
Because the χ2 value is much greater than the p = 0.05 value for 3 degrees of
freedom, it is accepted that the difference is NOT due to chance variation in the data.
e. Explain the difference between the expected and observed outcomes.
The genes for plant height and kernel color are linked – that is, they are on
the same chromosome and are not independently assorted.
4. In an experiment, yeast cells are introduced into a flask of growth media. A sample of the
media is removed at intervals, and the
number of yeast cells are counted. A graph
of the total number of yeast cells vs. time is
given at right.
a. At what time is the rate of growth the
greatest?
t = 7 hrs
b. Calculate the average growth rate for the
time interval from 5 hrs to 10 hrs. Include an
appropriate unit with your response.
Growth Rate =
25000 - 3000
= 4400 cells/hr
10 hrs - 5 hrs
c. If the death rate of yeast cells during the interval from 10 hrs – 12 hrs is 6000 cells/hr, what
is the birth rate during this interval?
Since the net growth rate is 0/hr, the birth rate = death rate = 6000 cells/hr
d. Explain the shape of the graph during the following time intervals:
i. From 0 hrs to 5 hrs.
Plentiful resources  exponential growth
ii. From 10 hrs to 12 hrs.
Population at carrying capacity  no growth
iii. From 12 hrs to 15 hrs.
Resources exhausted, or toxic byproducts built up  population collapse
5. Stomata are structures in leaves of plants that facilitate the exchange of gases (oxygen and
carbon dioxide), while regulating the loss of water through transpiration. The density of stomata
in leaves of a species exhibits variation, and is genetically determined.
In an investigation to determine variation in stomata density in oak trees, a student counted the
density of stomata (number of stomata per cm2) in eight plants from each of three populations.
The results are provided in the table below:
Population
I
II
III
Plant
1
113
93
129
Plant
2
108
67
154
Plant
3
140
54
137
Plant
4
122
117
118
Plant
5
120
146
130
Plant
6
115
121
118
Plant
7
109
140
133
Plant
8
116
156
134
Standard
Error of
the
Mean
Mean
118
4
112
13
132
4
(a)
On the axes provided, create an appropriately labeled graph to illustrate the sample
means of the three populations to within 95% confidence (i.e. sample means ± 2 SEM).
(c)
Describe the independent and
dependent variables and a control treatment
for an experiment to test the hypothesis that
higher stomata density on oak leaves is
selected for in regions of bright sunlight.
Identify an appropriate duration of the
experiment to ensure that natural selection is
measured, and predict the experimental
results that would support the hypothesis.
160
Stomata Density (stomata/mm2)
(b)
Based on the sample means and
standard errors of the means, identify the
two populations that are most likely to have
statistically significant differences in mean
leaf stomata densities. Justify your
response.
140
120
100
80
Pop I
Pop II
Pop III