Lab 1 Artificial Selection
The purpose of a particular investigation was to see the effects of varying salt concentrations of nutrient
agar and its effect on colony formation. Below are the results
Determine the mean for each treatment and graph the results.
Determine the standard deviation for the treatment with 0% salt and 3% salt and then perform a t-test
to determine if the difference between the means of these two trials is significant.
No Treatment
3% Salt
n
 ( X i  X )2
S=
n 1
n

286
 8.5
4
(X
S=
i
 X )2
n 1

39
 3.1
4
Predict how many colonies would be on an agar plate that contained 9% salt and the colony originated
from the plate with 9% salt.
There should be considerably more as the colonies are salt tolerant.
Explain how this is an example of artificial selection
Certain bacteria possessed the trait for salt tolerance and man manipulated the environment. Those
cells that possessed the trait are selected for or have an advantage.
Explain how this may happen as an example of natural selection.
Certain bacteria possessed the trait for salt tolerance. Something in the environment changes so that it
is more saline. Those cells that possessed the trait for salt tolerance are selected for or have an
advantage.
.
Lab 2 Hardy-Weinberg
1. The frequency of two alleles in a gene pool is 0.1 9(A) and 0 .81(a). What is the percentage in the
population of heterozygous individuals? What is the percentage of homozygous recessives? Assume
that the population is in Hardy-Weinberg equilibrium.
A =.19 thus p=.19
a=.81 thus q=.81
According to Hardy-Weinberg Equilibrium,
Homozygotes recessives are = q2
Heterozygotes are =2pq
Heterozygotes= 2pq Or =2 x 0.19 x 0.81 =0.31
Homozygotes recessives= q2 Or 0.81 x 0.81 =0.66
2. An allele W, for white wool, is dominant over allele w, for black wool. In a sample of 900 sheep, 891
are white and 9 are black. Estimate the allelic frequencies in this sample, assuming that the population is
in equilibrium.
The percent of homozygotes recessives is =
9
Or 0.01 and thus q2 =0.01
900
q is the allelic frequency for allele w, and p is the allele frequency for allele W.
q=
q2
q=
0.01 Or 0.1
p + q = 1 thus p = 1 – q Or p = 1 – 0.1 Or 0.9
p=0.9 and q=0.1 Or W=0.9 and w=0.1
3 In a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygote
genotype of a certain trait is 0.09. What is the percentage of individuals homozygous for the dominant
allele?
What has been given is the frequency of the recessive homozygote genotype which is q2 . What is asked
for is p2. Both p and q need to be determined
q=
q2
q=
0.09 Or 0.3
p + q = 1 thus p = 1 – q Or p = 1 – 0.3 Or p =0.7
p2 =(0.7)2 Or p2.= 0.49 X 100 (for answer in a percentage)
p2 =49%
4 In a population that is in Hardy-Weinberg equilibrium, 38 % of the individuals are recessive
homozygotes for a certain trait. In a population of 14,500, how many of the individuals will be
homozygous dominant individuals, and heterozygous individuals.
The problem gives the frequency of the recessive homozygote genotype which is q2 . You need to
determine the value for p2, and 2pq, and then use those frequencies to determine the number of
individuals in the population with that genotype. Both p and q need to be determined
q=
q2
q=
0.38 OR q= 0.616
p + q = 1 thus p = 1 – q Or p = 1 – 0.616 OR p =0.384
Number of individuals that are homozygous dominant=p2 (14,500) Or (0.384)2 (14,500) =2,138 (rounding
because there cannot be a fraction of a person).
Number of individuals that are heterozygous =2pq x (14,500) Or 2 x 0.616 x 0.384 x 14,500=6,860
5 Allele T, for the ability to taste a particular chemical, is dominant over allele t, for the inability to taste
it. At a university, out of 400 surveyed students, 64 were found to be nontasters. What is the percentage
of heterozygous students? Assume that the population is in equilibrium.
The percent of homozygotes recessives is =
64
Or 0.16 and thus q2 =0.16
400
q is the allelic frequency for allele t, and p is the allele frequency for allele T
q=
q2
q=
.16 Or 0.4
p + q = 1 thus p = 1 – q Or p = 1 – 0.4 =0.6
p=0.6 and q=0.4 Or T=0.6 and t=0.4
The heterozygote students is equal to 2pq Or Tt= 2 x 0.6 x 0.4 Or Tt= 0.48 X 100 (for a percentage) 48%
of the population is heterozygous
6 In humans, Rh-positive individuals have the Rh antigen on their red blood cells, while Rh-negative
individuals do not. Assume that a dominant gene Rh produces the Rh-positive phenotype, and the Rhnegative phenotype produces by its recessive allele rh. In a population that is in Hardy-Weinberg
equilibrium, if 160 out of 200 individuals are Rh-positive, what are the frequencies of the Rh allele and
the rh allele at this locus?
The problem gives the number of individuals with the dominant phenotype which can be used to
determine the frequency of the recessive phenotype Or genotypes which is q2. q2 can be used to
determine q and p.
The number of individuals that are rh-negative= 200-160 (number of Rh-positives) = 40 individuals
The frequency of this genotype is
q=
q2
q=
40
Or = 0.2 are rh-negative Or q2
200
.2 OR q= 0.45
p + q = 1 thus p = 1 – q Or p = 1 – 0.45 OR p =0.55
Rh-allele= 0.55 and rh allele=0.45
7 In corn, yellow kernel color is governed by a dominant allele for white color W and, by its recessive
allele, w. A random sample of 100 kernels from a population that is in equilibrium reveals that 9 are
yellow and 91 are white. What are the frequencies of the yellow and white alleles in this population?
What is the percentage of heterozygotes in this population?
The problem gives the number of individuals with the dominant phenotype and recessive phenotypes.
It is the recessive phenotype that can be used to determine the frequency of the recessive phenotype or
genotypes which is q2. q2 can be used to determine q and p.
The frequency of this recessive genotype is
q=
q2
q=
9
Or = 0.09 are ww Or q2
100
.09 Or q= 0.3
p + q = 1 thus p = 1 – q Or p = 1 – 0.3 OR p =0.7
W= 0.7 and w =0.3
The percentage that is heterozygous is= 2pq x100 Or 2 x 0.3 x 0.7 x100= 42%
8 A rare disease which is due to a recessive allele (a) that is lethal when homozygous (aa), occurs with a
frequency of one in a million. How many individuals in a town of 14,000 can be expected to carry this
allele?
The problem gives the number of individuals with the recessive phenotypes. It is the recessive
phenotype that can be used to determine the frequency of the recessive phenotype or genotypes which
is q2. q2 can be used to determine q and p.
The frequency of this recessive genotype is
q=
q2
q=
1
Or = 0.000001 is the frequency of aa Or q2
1, 000, 000
0.000001 OR q= 0.001
p + q = 1 thus p = 1 – q Or p = 1 – 0.001 Or p =0.999
A= 0.999 and a =0.001
The number of individual in a population that is heterozygous is= 2pq x14000 Or
2 x 0.999 x 0.001 x14000= 27.97 and rounding gives us 28 individuals are carriers Or Aa
Lab 3 Comparing DNA sequences
Below are strands of DNA from 4 different species. Base on the DNA sequence, answer the following
questions.
I. A C C G G T T A A A C A T T A G G G A C C T T A T G G A A A A C T C A C G A G C C C G G A T T A G G C
II . A C C G G T T T A A C A T T A G G C A C C T T A T G G G A A A C T C A T G A G C C C G G A T T A G G C
III. A C C G G T TG A A C A T T A G G C A C C T T A T G G G A A A C T C A T G A G C C C G G A T T A G G G
IV. A T C G G T C G A A C A T T A G…… A C C T T A T G G G A A A C T C A T G A G C G C G C A T T A G G G
1. Which two species are most closely related and why?
Species II and III seem to be most closely related as they have the fewest differences in their DNA
sequences.
2. Which species is more distantly related to the other species and why?
Species IV because there are two nucleotides that have been deleted and also has the greatest
differences in their DNA sequences.
3. What sorts of mutations are found in the DNA of species # 2 when compared to species #3? How will
that affect the resulting protein?
These are point mutations and may result in an amino acid substitution.
4. What sorts of mutation are found in the DNA of species #3 when compared to the species #4? How
will that affect the resulting protein?
There are point mutations and may result in an amino acid substitution, however what is even more
detrimental is the deletion of two nucleotides which will result in the protein from that point being
changed.
Lab 4 Osmosis
A student was working with potato cores and sucrose solutions with varying molarities. The molarities
of the solutions were 0, 0.2,0 .4, 0.6, 0.8, and 1.0. The student measured the initial mass of the potato
cores and then after 24 hour the student measures the cores again. The student then realizes that the
molarities of the beakers were never labeled. Based on the percent change mass of the potato cores
determine the molarity of each solution. Also determine the water potential of each solution.
Graph the percent change in mass versus the molarities of the solutions and determine the molarity
where the potatoes were isostonic.________________________________________________________
Now determine the water potential of the potatoes
Lab 5 Photosynthesis
An experiment involved plant tissue disks submerged in a solution of water and sodium bicarbonate.
The time it took for the disk to float to the top of the solution was recorded. Graph the data below and
determine the median time it took for 5 disc to rise to the top or ET 50 .
This experiment was repeated four times varying the light intensity of the light. The data was recorded
below . Determine 1/ET50. Graph the light intensity versus 1/ET50.
Does this relationship seem to be proportional?
The correlation is .95 which appears to be proportional
Draw in a line if this experiment were repeated under varying light intensities of green light. Justify your
how you drew your line.
Very little photosynthesis takes place in green light. The amount that does is due to the carotenoids and
xanthophylls.
Design an experiment that to investigate another factor of photosynthesis and the protocol you would
use and also justify the results you would expect.
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
Lab 6
Below is data presented from an investigation comparing the oxygen consumption of germinating pea at
25o C versus germinating peas at 10o C. Respirometers with KOH inside were used in measuring the
amount of oxygen consumed. In addition there was a control used which was a respirometer that
contained only beads and no germinating peas. The data table is only partially completed. Complete
the data table and graph the corrected difference versus time.
1. What is the effect of temperature on the rate of cellular respiration?
It is a proportional relationship. If the time that the peas are allowed to respire doubles, then the
amount of oxygen the peas consumes will double
2. What is it important to have a respirometer filled with beads?
It is a control to account for any changes in the volume of gas due to changes in the environment such
as temperature or pressure.
3. What is the purpose of KOH?
The purpose of the KOH is to remove the CO2 from the air.
4. Determine the average rate of respiration for germinating peas at 10oC and 25oC.
total # oxygen consumed
total # time
0.24 mL
mL
Rate of oxygen consumption mean rm temp=
=0.12
20 min
min
0.12 mL
mL
Rate of oxygen consumption mean cold temp=
=0.05
20 min
min
Rate of oxygen consumption mean=
The experiment was repeated four more times and the data is collected below.
Determine the standard deviation of each set of data and perform statistical with a t-test to determine if
the difference between means of the trials are statistically significant different.
*Remember that the t-Test will not be tested on the AP exam but standard deviation is fair game. In the
example above 19.04 is greater than 2.306 with 8 degrees of freedom. The null hypothesis is rejected
and that data in this experiment supports the hypothesis that a decrease in temperature decreases the
rate of respiration.
Using the means from the two trials determine the value Q10 .
 K2 
Q10  

 K1 
(
10
)
( t 2 t1)
(
10
)
 0.116  (2515)
=
=1.75 This means that the reaction rate almost doubles when

 .005 
the temperature is increased 10o C.
Lab 7 Mitosis
Above is an onion root tip observed under a microscope. In the data table shown below, count the
number of cells in each phase of mitosis. Combine prophase and prometaphase.
Determine the percentage of cells spent in interphase versus mitosis.
If the life cycle of the onion root tip cell is 24 hours, how many minutes will the average cell spend in
phase of mitosis
This is a root tip that has been treated with a 1 M concentration of caffeine. Determine the percentage
of cell involved in mitosis.
Does it appear that caffeine has an effect on the number of cells involved in mitosis?
Yes, the percentage of cells involved in mitosis decreases almost by half.
Do a Chi-square analysis that determines whether a 1 M concentration of caffeine significantly affects
mitosis.
Percentage in mitosis untreated = 24%
Percentage in interphase untreated=76%
Expected in the caffeine treatment
Mitosis = .24 x 83 cells= 20
Interphase = .76 X 83 cells= 63
Observed
Expected
Difference
Chi square calculation
X2  
Mitosis
11
20
-9
Interphase
72
63
9
(o  e)
81
81
= ( ) +(
)= 5.34
e
20
63
With one degree of freedom, a value of 5.34 is greater than 3. 84. So there is a high probability that the
difference between the treated and nontreated root tips is significant.
Is this value significant?
Part III.
The roots of several onion bulbs were removed. Then the remaining severed root mass of onions bulbs
were submerged in a beaker of distilled water, and others were submerged in a beaker of water seeped
with weeping willow bark. It has been reported that a substance found in the bark of the weeping
willow is a root stimulator. The roots were allowed to regenerate for five days. After five days the roots
were measured. To determine if the there is a substance in the weeping willow bark that is a root
stimulator, statistical analysis must be done. Part of the statistical analysis would include a t-test and
determination of the standard deviation for each set of data. Determine the standard deviation for each
set of data. Perform a t-test to determine if the differences between the two sets of data are
significantly different in determining if a substance found in the bark of the weeping willow is a root
stimulator.
Untreated
Treated
n
(X
S=
i
n
 X )2
n 1

21.98
 2.34
4
(X
S=
i
 X )2
n 1

29.73
 2.73
4
*Remember that the t-Test will not be tested on the AP exam but standard deviation is fair game. In the
example above, the t-test value of 4.8994 greater than 2.306 with 8 degrees of freedom and therefore,
the null hypothesis is rejected. The data in this experiment supports the hypothesis that a there is a
substance in the bark of the willow tree that stimulates root growth and mitosis.
Lab 8
A scientist is using an amplicillin-sensitive strain of bacteria that cannot use lactose because it
has a nonfunctional gene in the lac operation. She has two plasmids. One contains a
functional copy of the affected gene of the lac operon, and the other contains the gene for
amplicillin resistance. Using restriction enzymes and DNA ligase, she forms a recombinant
plasmid containing both genes. She then adds a high concentration of the plasmid to a tube of
the bacteria in a medium for bacterial growth that contains glucose as the only energy source.
This tube (+) and a control tube (-) with similar bacteria but no plasmid are both incubated under
the appropriate conditions for growth and plasmid uptake. The scientist then spreads a sample
of each bacterial culture (+ and -) on each of the three types of plates indicated below.
1. If no new mutations occur, it would be most reasonable to expect bacterial growth on which of
the following plates and be sure to justify your answer.
Plate 1 Yes there will be growth because there is no antibiotic to kill the bacteria.
Plate 2 Yes there will be growth because the some of the bacteria has taken up the plasmid and
is now resistant to amplicilin.
Plate 3 Yes there will be growth because the some of the bacteria has taken up the plasmid and
is now resistant to amplicilin and can also use lactose an energy source.
Plate 4 Yes there will be growth because there is no antibiotic to kill the bacteria.
Plate 5 No there will be no growth because there is antibiotic to kill the bacteria and it does not
have a plasmid to resist the antibiotic.
Plate 6 No there will be no growth because there is antibiotic to kill the bacteria and it does not
have a plasmid to resist the antibiotic.
2. Why were restriction enzymes used in this experiment?
The same restriction enzymes were used to cut the plamid and the genes in order to make sure
that the plasmid will take up the gene. They will have complementary “sticky” ends.
3. If DNA ligase was not used during the preparation of the recombinant plasmid, bacterial
growth would most likely have occurred on which plates and justify your answer.
The plasmid would not be functional and therefore there would not be any growth on plates
2,3,5 or 6.
4. If another experiment was done with the cultures as shown above, using medium that
contained lactose as the only energy source, growth would most likely occur on which of the
following plates?
Growth would occur with plates 7, and 8 but not plates 9 and 10 because the bacteria cannot
use lactose as an energy source because they do not have the plasmid on plates 9 and 10.
Lab 9
The diagram below shows a segment of DNA with a total length of 4900 base pairs. The arrows indicate
reaction sites for two restriction enzymes (enzyme X and enzyme Y).
1. Explain how the principles of gel electrophoresis allow for the separation of DNA fragments.
DNA has an overall (-) and it is inserted in a wells on a gel made of aragose. Agarose is porous and DNA
fragments can move through the gel. The size of pores can be regulated by adjusting the concentration
of agarose used. The gel is inserted into an electrophoresis box and covered with a buffer. This box has
poles so that when connected to direct current one is positive and one end is negative. The gel is
inserted so that the top of the gel is closest to the negative pole. The box is then connected to direct
current. The negative pieces of DNA start to migrate to the positive pole. The smaller DNA fragments will
migrate at a faster rate than larger fragment.
2. Describe result you would expect from the electrophoretic separation of fragment from the following
treatment of the DNA segment above. Assume that the digestions occurred under appropriate
conditions and went to completion.
I.DNA digested with only enzyme X It would separate into four distinct bands on the gel.
II.DNA digested with only enzyme Y. It would separate into two distinct bands on the gel.
III. DNA digested with enzyme X and Y. It would separate into five distinct bands on the gel.
IV. Undigested DNA. It would separate into one distinct band on the gel.
3. Explain both of the following. The mechanism of action of restriction enzymes, and the different
result you would expect if a mutation occurred at the recognition site for enzyme Y.
Restriction enzymes cleave pieces of DNA at specific DNA sites or DNA sequences. Sometimes it
results in sticky ends or blunt ends. Each restriction enzyme recognizes different DNA sequences. If
there were a mutation at Y, then the restriction enzyme would not recognize that sequence and
therefore would not cleave the DNA.
One female and two male adult hawks nested together. All the birds hunted in group and caught prey to
feed the three nestlings. The female spend the most time I the nest, but allowed Male C to help care for
the nestlings. Male C would chase male B away. Prove who the parents are and how DNA technology
aided in this quest.
The unique bands shows that male B is the parent for nestling #1 and that male C is the parent for
nestling #2.
Lab 11
1. Explain the results of the lab in terms of water potential.
The lower the water potential, the faster the rate of transpiration.
2. Why was it important to divide the water loss by the surface area of the leaves?
Each plant has a different surface area and the surface area should be proportional to the rate of
transpiration.
The experiment was repeated four more times for the untreated plant and the fan. The data is collected
below.
Untreated
Treated
n
 ( X i  X )2
S=
n 1
n
8.61

 1.47
4
(X
S=
i
 X )2
n 1

3.37
 0.92
4
*Remember that the t-Test will not be tested on the AP exam but standard deviation is fair game. In the
example above, the t-test value of 5.31 greater than 2.306 with 8 degrees of freedom and therefore, the
null hypothesis is rejected. The data in this experiment supports the hypothesis that there is a
significant difference in the transpiration rate of a plant with no factors and a plant that is experiencing
the wind or fan.
Lab 12
The activities of organisms change at regular time intervals. These changes are called biological rhythms.
The graph depicts the activity cycle over a 48-hour period for a fictional group of mammals called
pointy-eared bombats, found on an isolated island in the temperate zone.
(a) Describe the cycle of activity for the bombats. Discuss how three of the following factors might
affect the physiology and/or behavior of the bombats to result in this pattern of activity.
_ temperature
_ food availability
_ presence of predators
_ social behavior
(b) Propose a hypothesis regarding the effect of light on the cycle of activity in bombats. Describe a
controlled experiment that could be performed to test this hypothesis, and the results you would
expect.
Description of the cycle of activity (1 point)
A student could earn a point if he/she accurately summarized the graph. A simplistic statement such as,
“Bombats are active during the day and quiet at night” which ignored the shape and obvious peaks and
valleys of the graph did not receive the point. To earn this point, the student had to identify the peak of
activity at “noon,” “midday,” or “12:00 p.m.,” AND indicate a lower activity at “night.” The student
could also be specific about the lowest activity being at “midnight,” or “12:00 a.m.” The description had
to be clearly distinguishable from the rest of the answer and not simply implied in another part of the
response.
Discussion of how THREE factors might affect the physiology and/or behavior resulting in the
cyclic activity pattern (1 point each)
To earn points here, each of the descriptions had to (a) be biologically plausible and consistent with
typical mammalian behavior and physiology (no fictional biology); (b) indicate a cause and effect
relationship beyond a simple restatement of the question. This had to include at least a very brief
indication of how or why the factor had any effect at all on the bombat — or in some cases its prey; and
finally, the discussion could not be inconsistent with the part of the curve described or time of day
referenced in the explanation.
Elaboration on any one of the three factors (1 point)
Here the readers were looking for exemplary descriptions of physiology and/or behavior that reflected an
unusual depth of understanding and clarity of expression. With special regard to temperature, a student
who demonstrated an understanding that the activity curve was different from a temperature curve or that
mammal physiology, unlike that of ectotherms, is typically insensitive to temperature could earn an
elaboration point.
2. (b) A maximum of 7 points
Hypothesis (1 point)
The student was required to indicate that a CHANGE in the light (intensity, duration, wavelength) causes
a CHANGE in the cycle of activity or biorhythm. There also could be a prediction of a change of light
having no effect on the cycle of activity. Like the description of the curve above, the hypothesis statement
had to be clearly distinguishable from the rest of the answer and not simply implied in another part of the
response. The student may have failed to earn this point if the experiment he/she designed below clearly
used a different light characteristic (independent variable), and/or produces a different result (dependent
variable) from the ones indicated in his/her hypothesis.
_____________________________________________________________________________________
Part II
Forty flies were put into a choice chamber with two chambers. In one chamber there was a cotton
ball soak with vinegar. The other chamber had nothing. After 20 minutes the number of flies were
counted in both chambers. This was repeated four more times. Perform a chi-square analysis to
determine if the difference between in the number of flies found in the two chambers is significant.
Chi square calculation X 2 

(o  e)
24
76
=(
) +(
)= 11.52
e
100
100
With one degree of freedom, a value 11.52 is greater than 3. 84. So there is a high probability that the
difference between the treated and nontreated root tips is significant.
Lab 13 Enzyme Lab
An investigation was determining the amount of oxygen released from the decomposition of hydrogen
peroxide with enzyme catalase. This investigation was conducted at two different temperatures. Graph
the data presented in the data table.
1. Is the relationship between time and amount of oxygen released proportional? Explain the
relationship. No this is not a linear relationship, and this is not proportional relationship. The
reaction is much faster in the beginning of the reaction and then it continues to slow down.
2. During what time frame is the rate of the reaction the fastest? Explain why. It is fastest at the
beginning of the reaction. This is because the concentration of the hydrogen peroxide is
greatest at the beginning of the reaction. This increases the likelihood that catalase will interact
with the hydrogen peroxide. As the reaction proceeds the concentration of the hydrogen
peroxide decreases and as a result there is a decrease in the reaction rate.
3. What is the effect of a decrease in temperature in this investigation? A decrease in temperature
will cause a decrease in the reaction rate. The molecules are moving slower, this decreases the
reaction rate as it is less likely that a collision will occur between the molecules.
4. Determine the rate of the reaction from 0-1 minute. At 25oC the reaction rate is 21 mL/min and
at 10oC is 8 mL/min.
5. Draw in a line for what the reaction would be if done at
37oC. Justify your prediction. The line should be above the 25oC
and flatten out at 80 mL as that is how much oxygen is present.
The molecules are moving at a faster rate so the reaction will also
be moving faster.
22 C
10 C
n
(X
S=
i
n
 X )2
n 1

(X
22.8
 2.39
4
S=
i
 X )2
n 1

9
 1.58
4
Determine the standard deviation of each set of data and perform statistical with a t-test to determine if
the differences between means of the trials are significantly different.
*Remember that the t-Test will not be tested on the AP exam but standard deviation is fair game. In the
example above, the t-test value of 9.21 greater than 2.306 with 8 degrees of freedom and therefore, the
null hypothesis is rejected. The data in this experiment supports the hypothesis that there is a
significant difference in the enzymatic decomposition of hydrogen peroxide at 22 oC and enzymatic
decomposition of hydrogen peroxide at 10 oC
Using the means from the two trials determine the value Q10 .
 K2 
Q10  

 K1 
(
10
)
( t 2 t1)
(
10
)
 20.8  (2215)
=
=2.01 This means that the reaction rate doubles when the

 9 
temperature is increased 10o C.