Mitosis and Meiosis LAB
Group Members:
Lab Station:
Standard: AP Big Idea #3 and SB2ce
Essential Question: How do eukaryotic cells divide to produce genetically identical cells or to produce gametes with half the
normal DNA?
Opening: Compare and contrast word wall activity for mitosis and meiosis.
Background:
Reproduction in living things can be achieved through either asexual or sexual processes. In asexual reproduction, offspring organisms
are genetically similar to the parent organism with limited, if any, introduction of DNA from another individual. In sexual reproduction,
offspring organisms are generated by combining one half of the DNA from each of two parent individuals to produce a diploid (2N)
offspring that has two complete sets of chromosomes and genes. The process of sexual reproduction enables greater genetic variation
in offspring than asexual reproduction. Multicellular, diploid, eukaryotic organisms use both types of cellular reproduction. Growth
and development of the organism occurs through asexual reproduction of cells through the process of mitosis where every normal
daughter cell is diploid, carrying two copies of each chromosome and set of genes. In mitotic reproduction, these daughter cells are
genetic replicates of both the parent cell and the sister cell. Most single cell eukaryotes reproduce by simple mitosis, producing more
replicates of the parent organism.
Mitosis
The mitotic cell cycle is a highly regulated process that insures accuracy in DNA replication and equal division into daughter cells.
Defects in accurate reproduction will cause abnormalities that will result in decreased viability of both the daughter cells and the
multicellular organism through a variety of mechanisms (including apoptosis and cancer). Mitosis is a continuous process that can be
described by a sequence of stages. The stages are defined by molecular and cytosolic events. Visual inspection of dividing cells
through the microscope has defined three main stages (Figure 1) that can be further subdivided (Figures 2, below figure 1, and figure
3 on the next page):
Interphase:
The cells look quiescent.
Mitosis:
The cells are separating DNA equally for two future daughter cells.
Cytokinesis:
The daughter cells are separated into two individual cells.
Interphase
In interphase, the cell may be arrested in this part of the cell cycle (referred to as the G0
part of interphase) as a diploid cell (2N). If the cell continues through the next mitotic cycle,
interphase only appears to be quiescent. This is when the cell grows in preparation of
another cycle (G1), DNA is replicated (S phase, cell becomes 4N), the DNA checked for errors
and repaired prior to mitosis (G2).
Mitosis is subdivided into four stages that can be identified
through the microscope: prophase, metaphase, anaphase,
and telophase.
Prophase
 The replicated DNA condenses around histone proteins and supercoils to form visible chromosomes.
• The nuclear membrane dissolves.
• Centrioles form and begin to migrate to opposite sides of the cell.
• Microtubules organize around the centrioles to form mitotic spindles
• Spindle fibers attach to the kinetochore proteins at the centromere (holding sister chromatids together).
Metaphase
• Centrioles complete migration to opposite poles of the cell
• Sister chromatids line up along the midline of the cell
Anaphase
• Spindles shorten, pulling sister chromatids apart to opposite poles of the cell
Telophase
• Chromosomes de-condense
• 2 nuclear envelopes form around the separated chromosomes (each nucleus is 2N again)
Cytokinesis
Cytokinesis generally follows and divides the cytosol and cellular membrane to yield two cells that are identical. In plant cells, small
vesicles move along microtubules to the mid line. These vesicles fuse to form a cell plate, which grows to form the cell wall that
separates the cells.
Control of the cell cycle
To maintain normal growth and development it is essential that cells divide only when and where needed. Therefore, the transition
between the stages of the cell cycle is a tightly controlled process. Levels of proteins called cyclins build up in the cell and bind to
cyclin dependent kinase, forming CDK complexes. These CDK molecules either add or remove phosphate groups from substrates to
move the cell to the next stage of the cycle. In order to progress, however, the cell must pass through several “checkpoints.” These
checkpoints assure that cells only divide when needed and when DNA duplication is completed and without errors.
The three main checkpoints are:
Restriction Point – this occurs in
G1. After division the daughter
cells will continue to grow in
size for a short time but are
halted at the restriction point.
Here cells will either terminally
differentiate and enter what is
termed G0, or the cells will be
stimulated a by growth factor
and continue through the cycle
again.
G2/M Checkpoint - occurs
between the end of G2 and the
beginning of mitosis. Specialized
molecules read the newly
formed DNA and will delay the cell from entering mitosis if there are strand breaks or if inappropriate nucleotides are
incorporated. The cell cycle will only continue if repairs can be made, otherwise, the cell will die without completing the
cycle.
Metaphase/Anaphase Checkpoint - early in mitosis. Specialized proteins on each centromere, called the kinetochores,
activate if microtubules are attached and appropriate force is being applied. Activation allows the chromosomes to
separate. If activation does not occur, mitosis will halt and the cell will die without completing the cell
cycle.
Mutations in cells that lead to the loss of cell cycle control result
in a loss of control over growth, cell differentiation, and death.
Mutations may accumulate over time causing the cell clones to
become a tumor; the tumor may later become metastatic if the
cell acquires the ability to establish separate masses in distant
tissues. These cancerous mutations usually affect genes which
encode proteins involved in the control of other genes
(transcriptional regulation), cellular metabolism (speeding up or
slowing down cell growth), and/or the cell cycle of division and
growth. Also affected are hormones and their receptors,
regulatory molecules such as cytokines and their receptors, and
DNA repair mechanisms.
Meiosis
Sexual reproduction occurs through the production of gametes
which contain only one set of genes (haploid, 1N). When two
gametes fuse, a new diploid organism with a different
complement of gene copies than either parent is produced.
Haploid gametes are produced through the process of meiosis.
Meiosis consists of one round of DNA replication followed by two
nuclear divisions, meiosis I and meiosis II (See the figure to the
right). This results in the formation of four daughter cells, each
with only half the number of chromosomes of the parent. When two gametes combine during fertilization to form a
zygote, the diploid chromosome number is restored in the resulting organism. Typically, we think of gametes as cells
that come from either a male (sperm) or a female (ova) that are specialized to fuse with each other in species specific
way (mouse sperm cannot fertilize a whale ovum).
Under a microscope, mitosis and meiosis do not look very different. The phases of division are visually distinct and
named as their visual look-alikes in mitosis. However, the way replicated DNA is divided is very different in meiosis than
mitosis.
Prophase I:
The duplicated sister chromosomes condense
and are attached at a centromere as in mitosis.
Unlike mitosis, these sister pairs form a tetrad
with duplicated chromosome homologs from
both the maternally and paternally contributed
chromosomes. At this point DNA may cross over
between non-sister, chromatid homologs
forming unique chromatids that contain DNA
from both maternal and paternal
chromosomes.
Metaphase I:
The tetrads align at the center of the cell and
spindle microtubules attach to the
kinetochores.
Anaphase I:
The tetrads separate into two, duplicate
chromosomes that remain attached at the
centromere which move to separate poles of
the cell. Anaphase I is followed by telophase 1
and cytokinesis, resulting in two daughter cells
that each have 2N DNA. Mitosis II follows
immediately after meiosis I is completed
without further duplication of the DNA. The
single set of chromosomes (held together at the
centromere) are separated on the second spindle thus forming daughter cells that are now 1N. A starting diploid germ
cell has now formed 4 haploid gametes, each containing a mixture of DNA from both the mother and the father. This
method of division produces gametes that vary greatly between each other with respect to the allele combinations
available to pass to the next generation.
The example of meiosis that will be used in this investigation is Sordaria fimicola. S. fimicola is an ascomycete fungus
that is haploid for the bulk of its life cycle; the haploids comprise the individual fungal filaments, called hyphae, which
normally exist in a mass, called a mycelium, representing the “body” of the fungus, and the ascospores, from which
mycelia develop. The only diploid portion of the life cycle of S. fimicola occurs when the nuclei of specialized hyphae
come together.
These hyphae fuse to form a diploid zygote. This zygote then undergoes meiosis to produce the haploid ascospores,
yielding four haploid nuclei contained in a sac called an ascus. After meiosis I and II, the four haploid nuclei undergo
mitosis, resulting in an ascus containing eight haploid ascospores from one diploid zygote. Many asci form inside a
fruiting body called a perithecium.
Wild types sordaria are a dark brown, but we will grow this cultures in the presence of a tan mutant so that when
mycelium from a wild type fuses with the tan to form a zygote, we will be able to observe the results of crossing over
events that occurred as the zygote underwent meiotic division. Figure 6 shows a culture plate that was seeded with two
squares of wild type sordaria containing agar and two squares of tan type sordaria containing agar. The mycelium grow
out from these blocks so fusion will occur where the two types meet. This region of fusion is where you will find
perithecium containing ascospores from zygotes that came from a wild type and a tan parent (arrows). After meiosis I
and II, the haploid ascospores will be either tan or wild type. When the parathecium are gently ruptured, the ascospores
will be in strings of 8 that can easily be scored as tan or wild.
How these ascospores are arranged within the ascus is a direct representation of whether or not crossing over has
occurred between the centromere and the site for the gene for ascospore color. If no crossing over has occurred, the
ascospores will be arranged in a 4:4 manner. If crossing over has occurred, they will occur in a 2:4:2 or 2:2:2:2
manner(s).
Meisosis with crossing over
By observing the ascospore arrangement, the percentage of asci exhibiting crossover can be determined. This frequency
appears to be affected largely by the distance from the gene to the centromere. From the crossover frequency, the
distance in map units from the gene for ascospore color, and the chromosome centromere, can be calculated.
Pre-lab Questions:
1. Explain the difference between mitosis and binary fission.
2. Explain what crossing over is, when it occurs, and why it is important.
3. How is the cell cycle different for animals and fungi?
4. How do mutations affect the cell cycle?
5. When do checkpoints occur during the cell cycle? What are the purpose of checkpoints during the cell cycle?
6. Summarize how mitosis and meiosis different and similar?
7. Explain the various ways the cell cycle controlled?
8. How is cell division different for plant and animal cells?
Part 1 - Materials
1 bag of:
40 red, 40 pink, 4 clear centrioles, 4 yellow centromeres
Part 1 – Procedure (write your descriptions in the table provided at the end of the procedure section)
To simulate Interphase
1. Construct two strands of seven red pop beads and attach each strand to a yellow centromere. Repeat with two
strands of seven pink pop beads and a yellow centromere. These represent a homologous pair of chromosomes
(red from the father and pink from the mother).
2. Picture an imaginary boundary in the center of your desk. This boundary represents the nuclear membrane.
Place the chromosomes in the center of the imaginary nucleus.
3. DNA replication occurs, producing a duplicate of each chromosome. Construct two chromosomes identical to
the ones you made previously. Each of the duplicated chromosomes is called a chromatid. Join both red
chromatids at the centromere to form a pair of sister chromatids. Repeat this process for the pink chromosome.
4. Place a pair of plastic centrioles at the ninety degree angle, just outside of your nuclear membrane. The
centrioles also replicated during interphase, so place another pair next to them in your cell.
5. Illustrate your simulation of interphase and in your own words provide a brief description of what happens
during interphase.
To simulate Prophase
6. Move your two pairs of centrioles to opposite poles (sides) of the cell (your desk).
7. Illustrate your simulation of prophase and in your own words provide a brief description of what happens during
prophase.
To simulate Metaphase
8. Center your chromosomes along an imaginary metaphase plate with the centrioles still at the opposite poles of
the cells.
9. Illustrate your simulation of metaphase and in your own words provide a brief description of what happens
during metaphase.
To simulate Anaphase
10. Separate and move the centromeres of each chromosome toward opposite poles of the cell. Notice how the
arm of each chromosome trails the centromeres to the poles.
11. Illustrate your simulation of metaphase and in your own words provide a brief description of what happens
during anaphase.
To simulate Telophase and Cytokinesis
12. Move one red strand and one pink strand to the centrioles they were heading toward during anaphase. Imagine
a cleavage furrow developing between each nuclei and separating the cell into two daughter cells.
13. Note how each cell now contains one red and one pink chromosome, as well as one pair of centrioles, exactly
like the cell with which you began.
14. Illustrate your simulation of metaphase and in your own words provide a brief description of what happens
during telophase/cytokinesis.
Interphase
Illustration
Description
Prophase
Metaphase
Anaphase
Telophase/Cytokinesis
Part 2 – Procedure: Mitosis using prepared animal and plant slides
1. Obtain a compound microscope, the animal cells slide and the onion mitosis slide.
2. Observe the prepared microscope slide of onion root tip first at 100X (10X objective and 10X ocular) then at 400X (40X
objective, 10X ocular). Look for cells in mitosis. Depending on the quality of your microscope you may be able to
distinguish the various phases of mitosis. Use oil immersion if available.
3. Observe the prepared microscope slide of the animal mitosis first at 100X, then 400X. Look for cells in mitosis and
classify the stages. Compare and record similarities and contrast differences between the animal and plant cell mitosis.
4. Examine at least three fields of view of the apical meristem of the onion root tip at 400X. In each view, count and
record the number of cells in interphase and the various stages of mitosis.
5. Calculate the percentage of total cells counted in interphase and in each stage of mitosis. If the individual phases
cannot be discerned, calculate the percentage in interphase and in mitosis.
Table 1. Number of cells in each mitotic stage.
Stage
# of cells in
% total number
onion slide
of cells
Time (minutes in
each stage)
# of cells in
animal slide
% total number
of cells
Interphase
Prophase
Metaphase
Anaphase
Telophase
Assuming that it takes an average of 24 hours (1,440 minutes) for onion root tip cells to complete the cell cycle, calculate
the amount of time cells spend in each phase of the cycle. Use the formula provided below. Enter your results in Table 1.
% of cells in phase x 1,440 minutes = _______ minutes cell spent in phase
7. Make a pie chart representing the amount of time spent in each stage of mitosis. Title and label all portions of the pie
chart below.
Prophase is normally the longest phase of mitosis, due to the complexity of the events occurring during prophase. These
complex events take a relatively long time for the cell to perform:chromatin condensing and thickening, nuclear
membrane dissolving, and the early stages of spindle development.
Part 3 – Procedure: Root Tip Squash
1. Obtain some garlic root tip. Blot as much excess water from the root tips as possible. Any excess water on the slide
will affect your results. Do not allow the root tips to dry out, however.
2. Using a scalpel, cut off the end of one of the emergent root tips; the section should be approximately 1 to 2 mm long.
Place the root tip on a clean microscope slide and apply two or three drops of hydrochloric acid (HCl) to the root tip.
3. Holding the slide with a clothespin, pass it through the flame of a Bunsen burner for five seconds (five passes over the
flame). Pass the slide through the flame of the Bunsen burner. Do not hold the slide directly in or over the flame.
4. Without harming the root tip, blot the specimen with a paper towel to remove the excess HCl. You may wish to touch
a corner of the paper towel to the drop on the slide and allow the paper towel to soak it up. This may not remove the
liquid from the slide as effectively as blotting, but it will not disturb the root tip.
5. Add a few drops of carbol fuchsin stain, covering the root tip.
6. Pass the slide through the flame of the Bunsen burner approximately 10-15 times. Let the slide stand for one minute.
Pass the slide through the flame of the Bunsen burner. Do not hold the slide directly in or over the flame. Without
disturbing the specimen, use a paper towel to remove the excess stain.
7. Cover with a coverslip. Using a pencil eraser or other blunt instrument, gently press down on the coverslip to squash
and spread out the root tip. Blot off the excess stain, if any, that may have come out from under the coverslip.
8. Observe the slide under a microscope at 100X. Locate the apical meristem. Examine the slide at 400X. Locate cells in
the various stages of mitosis, and make sketches of what you find.
Table 2. Number of cells in each mitotic stage of the root tip.
Stage
# of cells in field # of cells in field # of cells in field
1 (directly at the 2 (middle of root 3 (opposite end
tip)
tip)
of root tip)
Interphase
Prophase
Metaphase
Anaphase
Telophase
Field of View Images for each magnification observed (10, 40, 100x)
Total # of cells
% total number
of cells
Part 4 – Application: Environmental Effects on Mitosis
Scientists reported that a fungal pathogen, may negatively affect the growth of soybeans (Glycine max). Soybean growth
decreased during three years of high rainfall, and the soybean roots were poorly developed. Close relatives of R. anaerobis are
plant pathogens and grow in the soil. A lectin-like protein was found in the soil around the soybean roots. This protein may
have been secreted by the fungus. Lectins compounds, such as phytohemagglutinin, induce mitosis in some root apical
meristem tissues. In many instances, rapid cell divisions weaken plant tissues. You have been asked to investigate whether the
fungal pathogen lectin affects the number of cells undergoing mitosis in a different plant, using root tips of onions.
Based on the information above create an experimental hypothesis (indicating whether you would predict onion roots to be
affected in a similar manner to the soybeans having increased mitotic divisions or not).
Experimental Hypothesis (Ha):
_______________________________________________________________________________________________________
_______________________________________________________________________________________________________
Now write the null hypothesis (Ho) (this is the opposite of your experimental hypothesis):
_______________________________________________________________________________________________________
_______________________________________________________________________________________________________
Root tips of onions were grown in both lectin exposed soil and soil will no lectin to determine the number of cells undergoing
mitosis. Using the experiment described, identify the following variables:
Independent variable:
Dependent variable:
Control:
The data below is representative of number of cells seen in interphase, which would look like a dark condensed region in a cell
and the combined number of cells in any of the phases of mitosis(PMAT).
Table 1. Table of observed (o) number of cells for the onions treated with lectin and without lectin.
Interphase
Mitosis
Total
Control – no lectin
148
25
173 (A)
Treated with lectin
161
88
249 (D)
Total
309 (B)
113 (C)
422 (N)
Below is a table for expected (e) values for onion root cells exposed to lectin and root cells not treat with lectin. You
must complete the table for all the expected values. You will need to use the equations provided while referencing the
variables indicated in table 1 ( A,B,C,D, and N)
Table 2. Table of expected (e) number of cells for the onions treated with lectin and without lectin.
Interphase
Mitosis
Control – no lectin
(A*B)/N =
(A*C)/N =
Treated with lectin
(D*B)/N =
(D*C)/N =
Table 3. Chi-Square Values for the onion cells treated with lectin and without lectin.
Group
Observed (o) value Expected (e) value
(o-e)
(o-e)2
(o-e)2 /e
Control Interphase
Control Mitosis
Treated Interphase
Treated Mitosis
Total of (o-e)2 /e = chi square (X2) = _________________
Chi-square is a statistical test commonly used to compare observed data with data we would expect to obtain according
to a specific hypothesis. For example, if, according to Mendel's laws, you expected 10 of 20 offspring from a cross to be
male and the actual observed number was 8 males, then you might want to know about the "goodness to fit" between
the observed and expected. Were the deviations (differences between observed and expected) the result of chance, or
were they due to other factors. How much deviation can occur before you, the investigator, must conclude that
something other than chance is at work, causing the observed to differ from the expected. The chi-square test is always
testing what scientists call the null hypothesis, which states that there is no significant difference between the expected
and observed result. That is, chi-square is the sum of the squared difference between observed (o) and the expected (e)
data divided by the expected data in all possible categories.
Once the Chi-square value is calculated the researcher needs to figure out if the value supports or rejects their null
hypothesis. Table 4, below, indicates the critical values in which you will compare your chi-squared value.
The degrees of freedom (df) equals the number of treatment groups minus one multiplied by the number of phase
groups minus one. In this case, there are two treatment groups (control, treated) and two phase groups (interphase,
mitosis); therefore df = (2-1)(2-1) =1.
The p value is 0.05, and the critical value is 3.84. If the calculated chi-square value is greater than or equal to this critical
value, then the null hypothesis is rejected. If the calculated chi-square value is less than this critical value, the null
hypothesis is not rejected.
Table 4. Critical Values of the Chi-Square Distribution
Probability
0.05
0.01
0.001
Analysis Questions:
1
3.84
6.64
10.8
2
5.99
9.21
13.8
Degrees of Freedom (df)
3
7.82
11.3
16.3
4
9.49
13.2
18.5
5
11.1
15.1
20.5
1. Based on the data is the null hypothesis accepted or rejected?
2. Explain what it means if your null hypothesis is rejected.
3. What other experiments should be performed to verify your findings?
4. Does an increased number of cells in mitosis mean that these cells are dividing faster than the cells in the roots
with a lower number of cells in mitosis?
Part 5 – Crossing Over in Sordaria
Refer to the introduction for information on the life cycle of Sordaria and crossing over.
Class sets of data and calculations for Sordaria crossing over is located in table 1.
A cross was made between wild type (+, black in color) and tan strains. The
resulting zygote produces either parental type asci, which have four black and
for tan spores in a row (4:4 pattern) or recombinant asci, which do not have this
pattern (2:2:2:2, 2:4:2). By studying the frequency of crossing over, you can
gather information that lets you draw a map of the relative location of genes on
a chromosome. A map unit is a relative measure of the distance between two linked genes, or between a gene and the centromere.
The greater the number of crossovers, the greater the map distance as seen in genes D-E and E-F in the diagram to the right. One
map unit equals one recombinant per 100 total events. The percentage of asci showing crossover divided by 2 equals the map units
in this activity. Each crossover produces two spores like the parents and two spores that are a result of the crossover. Thus, to
determine the number of crossovers, you must divide the number of asci counted by two since only half the spores in each ascus
result from crossing over.
Table 1. Number of Asci indicating crossing over events.
Number of Asci
Number of Asci
4:4 pattern
Crossing over
Total number of
Asci
%Asci showing
crossover /2
Gene to Centromere
distance
(Map Units)
425
51/2 =
25.5%
26
1st period
6th period
208
217
Analysis Questions:
1. Explain why we divide the percentage of asci showing crossing over (recombinant) by 2.
2. The published map distance between the gene and the centromere is 26 map units. How did the class data compare with
this distance?
3. What can account for disparities, if any, between the class data and the published data? If there were no disparities, why
might that be?
4. Can the same (or any) environmental factors you tested above affect the amount of crossing over that occurs in Sordaria?
How would you set up an experiment to test this? For example, how does humidity or pH affect the crossover frequency?
5. Knowing that there is a single gene that determines the spore color for Sordaria, use the data from your class to indicate
how far the gene is away from the centromere using the blank chromosome below.
6.
Calculate the map units between gene and centromere for the small sample of Sordaria asci given below.
Number of Asci Counted
Spore Arrangement
7
4 light/4 dark spores
8
4 dark/4 light spores
3
2 light/2 dark/2 light/2 dark spores
4
2 dark/2 light/2 dark/2 light spores
1
2 dark/4 light/2 dark spores
2
2 light/4 dark/2 light spores
7. Chi-squared value
Asci Configuration
Observed (o)
Expected (e)
((o-e)2)/2
4:4 wild:tan
208
Crossing over
217
totals
425
X2=
Create your hypothesis below. Anytime something is known, such as Mendels laws, that is the null hypothesis. By doing the
experiment you are trying to disprove the null or say that the law or theory is incorrect.
Remember when calculating the expected value you must multiply the expected map unit percent by 2 since you divide by 2
calculating the map unit value. In other words, 26x2=52 so .52x425= the expected value for crossing over.
H0- The map units for the Sordaria gene for color is 26 map units from the centromere.
Ha- ________________________________________________________________________________
8.
Does the data hold true to what scientists have come to accept for the gene distance to centromere? Explain why.
Part 6 – Loss of cell cycle control in cancer.
Background questions:
1. How are normal cells and cancer cells different from
each other?
2. What are the main causes of cancer?
3. What goes wrong during the cell cycle in cancer cells?
4. What makes some genes responsible for an increased
risk of certain cancers?
5. Do you think that the chromosomes might be different
between normal and cancer cells?
Many of us have family members who have or have had cancer.
Cancer can occur when cells lose control of their cell cycle and
divide abnormally. This happens when tumor suppressor genes,
such as p53 or Rb (retinoblastoma), are mutated.
The last background question is the focus of this part of the lab. With your group, form a hypothesis as to how the chromosomes of
a cancer cell might appear in comparison to a normal cell and how those differences are related to the behavior of the cancer cell.
For each of the following cases, look at pictures of the chromosomes (karyotype) from normal human cells. Compare them to
pictures of the chromosomes from cancer cells. For each case, count the number of chromosomes in each type of cell, and discuss
their appearance.
Case 1: HeLa Cells
HeLa cells are cervical cancer cells isolated from a woman named
Henrietta Lacks. Her cells have been cultured since 1951 and used
in numerous scientific experiments. Henrietta Lacks died from her
cancer not long after her cells were isolated. Lacks cancer cells
contain remnants of human papillomavirus (HPV), which we now
know increases the risk of cervical cancer. To the right is a
karyotype of a HeLa cell.
1. What differences do you observe between the normal
karyotype and that of the HeLa cell?
2. Explain how cancer cells could end up looking like this.
Case 2: Philadelphia Chromosomes
In normal cells, mitosis is usually blocked if there is DNA damage.
Sometimes, though, DNA damage makes cells divide more often.
Certain forms of leukemia have a unique feature called a
Philadelphia chromosome. Look at the karyotype of leukemia cells in
diagram to the right.
1. How do cells monitor DNA integrity?
2. How are the chromosomes in this cell, with leukemia, different
than that of chromosomes in normal cells?
3. How might these differences lead to cancer?