Math Grid-In Practice
AP Exam Review Fall 2014
Math Prompt #1
In a population of grasshoppers, the allele for tan color is dominant to the allele for green color. A drastic
increase in rainfall leads to selection against the tan phenotype. When the rainy season ends, 23% of the
remaining grasshoppers have the green phenotype. If this population is now in Hardy-Weinberg equilibrium,
what will the frequency of the tan allele be in the next generation? Calculate your answer to the nearest
hundredths.
Green phenotype = q2 = 0.23
Frequency of green allele = √.23 = 0.48
Frequency of tan allele (p) = 1 – 0.48 = 0.52
Math Prompt #2
The bacteria that cause pimples can be grown in the lab
using a suitable nutrient broth, where they will eventually
achieve exponential growth. Using the graph to the right,
calculate the mean rate of growth, in millions of bacteria per
hour, during their exponential phase. Calculate your answer
to the nearest hundredths.
Logarithmic growth takes place during the time where the slope is the greatest, approximately
between 12 and 30 hours. During that time (18 hours), the bacterial population started at 10
million and increased to 33 million (a difference in 23 million). Therefore, 23 million divided
by 18 hours gives a rate of 1.27 million bacteria per hour.
Math Prompt #3
What is the water potential for a sucrose solution in an open container that is 0.1M at standard room
temperature 23◦C? Calculate your answer to the nearest hundredths.
The solute potential is – (1) x (0.1M) x (0.0831 Bars/mole K) x (296 K) = - 2.46 Bars.
The pressure potential is zero because the solution is in an open container.
Therefore, (- 2.46) + 0 = -2.46
Math Prompt #4
Determine the SA to V ratio for a cube that has a side length of 2.5cm. Calculate your answer to the nearest
tenths.
SA = 6 x (2.5 cm x 2.5 xm) = 37.5 cm2
V = (2.5)3 = 15.6 cm3
SA/V = 37.5/15.6 = 2.4
Math Prompt #5
How many milliliters of a 0.5M glucose solution would you need in order to make 250mL of a 0.1M glucose
solution?
If CiVi = CfVf, then
(0.5M)(Vi) = (0.1M)(250mL)
Vi = 50mL
Math Prompt #6
Imagine that 20 people decide to start a new population, totally isolated from anyone else. Two of the
individuals are heterozygous for a recessive allele, which in homozygotes causes cystic fibrosis. Assuming this
population is in Hardy-Weinberg equilibrium, what fraction (expressed as a decimal) of people in this new
population will have cystic fibrosis? Calculate your answer to the nearest ten-thousands.
For this specific gene in this specific population, there are a total of 40 alleles, two of which are
the recessive cystic fibrosis allele (2/40 = 0.05 = q).
Since you need to be homozygous recessive to have cystic fibrosis, q x q = q2 = 0.052 = .0025
In other words, 25 out of 10,000 people (25%) will have cystic fibrosis.
Math Prompt #7
What is the pH of a solution with a hydrogen ion concentration of 1.33x10-8? Calculate your answer to the
nearest hundredths.
pH = -log (1.33 x 10-8) = 7.88
Math Prompt #8
A cell is at equilibrium with its surroundings. The molarity of the surrounding sucrose solution is 0.8M.
Calculate the solute potential of the surrounding sucrose solution. Assume standard room temperature of
23◦C. Calculate your answer to the nearest hundredths.
Solute Potential = -(1) (0.8M) (0.0831) (296) = -19.68 Bars
Math Prompt #9
A student has a 2g/L solution. He dilutes it and creates 3L of a 1g/L solution. How much of the original
solution did he dilute? Provide your answer to the nearest tenths.
If CiVi = CfVf
2Vi = 1(3)
Vi = 1.5
Math Prompt #10
The graph below shows the effect of an enzyme on activation energy. Plot A is the progress of the reaction
without an enzyme, and plot B is the same reaction with an enzyme. Determine the Gibbs free energy change
(∆G) for the reaction. Express your answer to the nearest whole number.
At height of curve:
18 kcal/mole – 14 kcal/mole = 4 kcal/mole decrease
-4 kcal/mole
Math Prompt #11
You and your friends have just measured the heights of your pets (in millimeters). The heights (at the
shoulders) are: 600mm, 170mm, 430mm, and 300mm. Calculate the standard deviation for the data collected.
Calculate your answer to the nearest hundredths.
Step 1: find the mean…
600+470+170+430+300 / 5 = 394
Step 2: find the standard deviation using the formula sheet…
Standard deviation is 147.32
Math Prompt #12
Suppose you were asked to investigate whether the fungal pathogen lectin affects the number of onion root tip
cells undergoing mitosis. You hypothesize that treating onion root tip cells with lectin will induce mitosis in the
cells. After designing and conducing your experiment, the following data are reported:
Interphase Mitosis
Control/Non-Treated 1550
200
Experimental/Treated 500
1250
Total
1750
1750
Calculate the chi-square value for this data. Report your answer to the nearest hundredth. If you cannot grid
this answer onto your sheet – just CIRCLE IT TO THE SIDE.
Remember – you need to determine the EXPECTED values using a table
such as this:
Solution:
First recognize that the data table given is observed data, and total the # of cells yourself…
Number of Cells
Group (observed)
Interphase
Control/Non-Treated
Experimental/Treated w/ Lectin
Mitosis
Total
1550
200
1750
500
1250
1750
2050
1450
3500
Now you need to determine the expected values to get a comparison…this is done with a simple addition of what you have from the observed data…
Number of Cells
Group (expected)
Control/Non-Treated
Experimental/Treated
w/ Lectin
Interphase
(1750)(2050)
3500
1025
(1750)( 2050)
3500
1025
Mitosis
(1750)(1450)
3500
725
(1750)(1450)
3500
725
Total
1750
1750
Now, let’s put together the Chi Square chart:
Control Inter
O
1550
E
1025
(O-E)
525
(O-E)2
275625
(O-E)2/E
268.90
Control Mitosis
Treated Inter
200
500
725
1025
-525
-525
275625
275625
380.17
268.90
Treated Mitosis
1250
725
525
275625
Chi Square: 1298.14
380.17
Math Prompt #13
Use the graph below to calculate the lag time in months between the change in the densities of the prey and
the predator populations. Give your answer to the nearest tenth of a month.
The correct answer can be any value from .8 to 2.
To calculate the lag time, calculate the differences between the prey and predator peaks or
valleys.