AP Exam: Math Practice
Name___________________________
1) A heterozygous red eyed female was crossed with a red eyed male. The results are shown below.
Red eyes are sex-linked dominant to white. Determine the chi square value for this scenario. Round to
the nearest hundredth. _5.23_ Also, identify the critical value that is valid for this scenario. _3.84_
Based on your chi square calculations, is your null hypothesis accepted or rejected? __rejected___
Phenotype
Red eyes
White eyes
# flies observed
136
64
2) Wisconsin Fast Plants have two very distinctive visible traits (stems and leaves). Each plant will either
have a purple (P) or green (p) stem and also have either have green (G) or yellow (g) leaves. Suppose
scientists carried out a cross that produced the following observed offspring: 240 Purple-stemmed,
Green-leaved plants, 63 Purple-stemmed, Yellow-leaved plants, 50 Green-stemmed, Green-leaved
plants, and 31 Green-stemmed, Yellow-leaved plants. Calculate the chi-square value for the hypothesis
that these offspring were the result of a dihybrid cross. 12.56
3) A census of birds nesting on a Galapagos Island revealed that 27 of them show a rare recessive
condition that affected beak formation. The other 63 birds in this population show no beak defect. If this
population is in HW equilibrium, what is the frequency of the dominant allele? Give your answer to the
nearest hundredth.
p = 0.45
4) In 2014 the Career Center High School student body was made up of 88% right handed students.
Being right handed (R) is the dominant trait over being left handed (r).
a. What is p and q for the population of 2014 Career Center High School students. p = 0.65 q = 0.35
b. Find the percent of the student body in 2014 that are homozygous right handed, heterozygous right
handed, and left handed. p2 = 42% 2pq = 46% q2 = 12%
5) The recombination frequency between gene A and gene B is 8.6% and the recombination frequency
between gene A and gene C is 7.8%. If the order of these genes on a chromosome is B-A-C, then the
approximate recombination frequency between gene B and gene C should be _16.4_%.
6) In Drosophila, there is a dominant gene for gray body color and another dominant gene for
normal wings. The recessive alleles of those two genes result in black body color and vestigial wings,
respectively. Flies homozygous for gray body and normal wings were crossed with flies that had
black bodies and vestigial wings. The F1 progeny were then test-crossed with the following results:
Phenotype
Gray body, normal wings
Black body, vestigial wings
Gray body, vestigial wings
Black body, normal wings
# Progeny
238
241
40
39
What is the recombination frequency for these two genes? _14.2_%
Would you say that these two genes are linked? _yes_ Why/why not? __R.F. < 50%______
_____________________________________________________________________________________
If they are linked, how many map units apart are they on the chromosome? _14.2 m.u._
7) Hydrogen peroxide is broken down to water and oxygen by the enzyme catalase. The following data
were taken over 5 minutes. What is the rate of enzymatic reaction in mL/min from 2 to 4 minutes?
Round to the nearest hundredth.
Time (min)
1
2
3
4
5
Amount of oxygen produced (mL)
2.3
3.6
4.2
5.5
5.9
0.95 mL/min
8) Calculate the probability (in fraction form) that two parents with the following genotype combination:
AaBBcc x AabbCc
will produce an offspring with the genotype combinations:
aaBbCc = 1/8
AabbCc = 0
9) The net annual primary productivity of a particular wetland ecosystem is found to be 7,550 kcal/m2.
If respiration by the aquatic producers is 13,400 kcal/m2 per year, what is the gross annual primary
productivity for this ecosystem, in kcal/m2 per year?
GPP = 20,950
10) Joe has 25 L of a 3 g/L solution. To this solution he adds 32 L of distilled water. What is the final
concentration of the solution?
1.3 g/L
11) There are 220 owls living in a certain forest at the beginning of 2014. Suppose that every year there
are 35 owls born and 22 owls that die.
a. What is the population growth rate? 13 owls/year
b. What is the maximum per capita growth rate of the owls over a year, assuming exponential growth
of the population? rmax = 0.06
12) Salem, Ohio has a population of 19,000 in the year 2014. The infrastructure of the city allows for a
carrying capacity of 24,000 people. rmax = .7 for Salem.
a. What will be the population growth rate for 2014 (include units)? 2,771 people/year
b. What will be the population size at the start of 2015? 21,771
13) The molar concentration of a sucrose solution in an open beaker has been determined to be 0.3M.
Calculate the solute potential at 27 degrees Celsius. Round your answer to the nearest tenth.
-7.5 MPa
14) A fertilized egg (zygote) undergoes 5 rounds of cleavage including cell partitioning. After the 5th
round, half of the cells divide again while the remaining cells undergo apoptosis. After this event those
remaining cells undergo 5 more rounds of cleavage. How many cells are present after these events?
_1,024__
2.0M sucrose
0.9M starch
0.7M NaCl
0.3M sucrose
1.2M starch
1.0M NaCl
The initial concentration of Side A and Side B are indicated above on either side of the U-tube.
The membrane shown is permeable to sucrose and NaCl but not starch.
15a) Initially,
which side is hypotonic? _B_
what is the molarity of the hypertonic side? _3.6 M_
15b) Please use the correct letters to fill in the blanks below.
In order to reach equilibrium:
sucrose will move from side _A_ to side _B_.
starch will move from side _---__ to side __---_.
NaCl will move from side _B_ to side _A__.
15c) After this system reaches equilibrium:
what is the molarity of each side? A = 3.05 M B = 3.05 M
which side will lose water? _A_
SIDE A
0.05M glucose,
0.02 M NaCl
SIDE B
0.04 M NaCl
The U-shaped tube in the diagram above is separated by a membrane that is impermeable to glucose
and NaCl but is permeable to water. Ψs = -2.24 MPa on Side A due to the 0.05M glucose solution and
0.02M NaCl solution contained within Side A of the U-tube. Ψs = -1.99 MPa on Side B due to the 0.04M
NaCl solution. As denoted by the arrows, the solution within Side A is experiencing a positive pressure,
while the solution within Side B is experiencing a negative pressure. Specifically, Ψp = +0.20 MPa on
Side A and Ψp = -0.02 MPa on Side B.
16a) Calculate the overall water potential (Ψ) within Side A of the U-tube. __-2.04 MPa__
16b) Calculate the overall water potential (Ψ) within Side B of the U-tube. __-2.01 MPa___
16c) Based upon these values, water will initially move from Side _B_ to Side _A_ of the U-tube via
osmosis.
17) The phosphorylation of glucose occurs via dehydration synthesis as follows:
glucose + Pi  glucose-6-phosphate + H2O
ΔG = +8.05 kcal/mole
The breakdown of ATP occurs via hydrolysis as follows:
ATP + H2O  ADP + Pi ΔG = -7.36 kcal/mole
Calculate the total ΔG of the reaction. Does the hydrolysis of ATP provide sufficient energy to power the
synthesis of glucose-6-phosphate? Why or why not?
ΔG = +0.69 kcal/mole
Insufficient energy…overall ΔG must be <0
18) You are conducting research on soil microbes that are involved with nitrogen fixation. Your primary
research has been done on one specific species of bacteria, but you’ve just discovered another bacterial
species that also may carry out nitrogen fixation. To initially compare this new species to the one that
you have been working with, you decide to do PCR to help you look for a familiar nitrogen-fixing gene
within the new species’ genome. You begin the PCR process and head home for the evening. Just as
you arrive home, the graduate student in your lab calls to say that an electrical breaker was tripped and
power went out in the lab, interrupting your PCR run. If you started with 178 molecules of DNA, how
many molecules of DNA should be in the PCR tube if the power went out after only 4 PCR cycles were
complete? __2,848__
19) When digesting a bacterial plasmid with the restriction enzyme BamHI and running the subsequent
fragment(s) on a gel electrophoresis apparatus, only one band appeared on the gel, corresponding with
a sequence length of 20 kilobase pairs. When digesting the same bacterial plasmid with a different
restriction enzyme, EcoRI, and running the fragment(s) on a gel, 3 bands appeared on the gel,
corresponding with sequence lengths of 12 kilobase pairs, 6 kilobase pairs, and 2 kilobase pairs,
respectively. When digesting the same bacterial plasmid simultaneously with both BamHI and EcoRI, 4
bands appeared on the gel, corresponding with sequence lengths of 8 kilobase pairs, 6 kilobase pairs, 4
kilobase pairs, and 2 kilobase pairs, respectively. Using this information, construct a map of this plasmid
below.