Answer Key
Unit 2 Genetic Processes
13. Daughter cells formed during cell division are
genetically identical to the parent cell.
Unit Preparation Questions
(Assessing Readiness)
14. In plant cells, a cell plate is formed during cytokinesis
and divides the cytoplasm into two. Animal cells are
divided by a contractile ring rather than a cell plate.
(Student textbook pages 154–7)
1. b
2. In both animal and plant cells: A–cell membrane
B–cytosol
C–mitochondrion
E–endoplasmic reticulum (smooth and rough)
G–Golgi apparatus
H–nucleus (or nuclear envelope)
L–nucleus (or nuclear envelope)
In plant cell only: I–central vacuole
J–cell wall (or membrane)
K–chloroplast
In animal cell only: D–ribosomes
F–vesicle
3. For efficient movement of materials the ratio of a cell’s
surface area to volume must stay within a range that
permits diffusion to all parts of the cell within a short
amount of time (to support the metabolic activity of
the cell).
4. d
5. Nucleus (contains) → chromosomes (composed of) →
DNA (which is made up of) → genes (which code for)
→ proteins
6. d
7. Students should draw a double helix and label any
small piece as a gene.
8. a
9. c
10. Mitosis involves the division of the nuclear material in
the nucleus and cytokinesis is the process that divides
the cytoplasm (including all organelles).
11. During cell division, DNA is tightly coiled in
chromosomes and looks “bunched.” The rest of the
time, the DNA looks more like long, loose threads.
12. A–centrosome
B–chromosomes
C–spindle fibres
D–centromere
15. A–interphase
B–growth and preparation (G1)
C–DNA replication
D–growth and preparation (G2)
E–mitosis
F–cytokinesis
16. Any two of: the DNA may be damaged; DNA
replication may not have occurred; there were not
enough nutrients to support cell growth; no additional
cells of that type are needed; cell death may have
been signalled
17. Sample answer: Skin cells have a short life span—skin
is easily damaged and must often be replaced. Nerve
cells have a long life span—these internal cells are
rarely damaged and, once an organism is full size, there
is no need to divide for growth.
18. Cancer results from uncontrolled cell reproduction and
the lack of cell death triggered by errors in DNA.
19. Sample answer: The digestive system is made of organs
including the stomach and small intestine, and tissues
like muscle (which move food along) or epithelium
(which line the tract). Muscle tissue is made of muscle
cells and epithelial tissue is made of epithelial cells.
Diagrams could look like the one that accompanies
question 38 on student textbook page 437.
20. Cell differentiation describes the process by which
a cell specializes to perform a specific function.
Because they perform different jobs, muscle cells
look very different from skin cells and nerve cells, for
example. If cells didn’t differentiate, they would all be
generalists—they would be able to do all functions,
but not very well.
21. Stem cells can specialize to become many types of cell.
Other types of adult cells can only divide to produce
cells identical to themselves.
Biology 11 Answer Key Unit 2 • MHR TR
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Chapter 4 Cell Division and
Reproduction
Learning Check Questions
(Student textbook page 164)
1. Interphase, mitosis, and cytokinesis
2. Interphase
3. At prophase, the cell’s chromatin condenses into
chromosomes. Each chromosome exists as two copies
of one chromosome, joined at a centromere.
4. When mitosis is inhibited, healing times increase.
5. Interphase would differ in length between cell types
because different cells have different functions that are
carried out during interphase. Mitosis and cytokinesis
are processes that would likely be consistent for all
cell types.
6. The daughter cells would either have twice as much
DNA as a healthy cell, or no DNA. Neither cell would
be viable. One would lack genetic material and the
other would have too much to be able to survive.
(Student textbook page 172)
7. A gamete is a haploid sex cell and a zygote is a
diploid cell.
8. Thirty-nine chromosomes; gametes are haploid,
meaning they contain half the number of
chromosomes that are in a diploid cell
9. See Figure 4.14 on student textbook page 172 for
illustration; each homologous chromosome still
consists of two sister chromatids. In mitosis, the sister
chromatids are separated during anaphase.
10. The gametes would have 46 chromosomes (23 pairs).
After fertilization, the zygote would have
92 chromosomes (four sets of 23).
11. The phases in meiosis II (metaphase II, anaphase II,
and telophase II) are most like the phases of mitosis
because, during these stages, the chromosomes
align on the equator of the cell, sister chromatids are
separated, and a new nucleus forms for each new cell.
12. During anaphase I and anaphase II, because this
is when the homologous chromosomes and sister
chromatids, are each separated
(Student textbook page 176)
13. The outcome of mitosis is genetically identical
offspring, whereas the outcome of meiosis is one
(or four) haploid cells with a genetically diverse
2
MHR TR • Biology 11 Answer Key Unit 2
representation of the parent cell. This leads to
genetically unique offspring.
14. 27 = 128 different gametes
15. Students may redraw the figure showing chromosomes
that are mixes of pieces of blue and yellow.
16. Because of independent assortment and crossing over,
all children will have some genetic material from each
of their ancestors.
17. Alleles that are found close together on the same
chromosome will be inherited together more often
than those that are either far apart on a chromosome or
on different chromosomes.
18. The ideal donor is an identical twin, because they are
genetically identical.
(Student textbook page 185)
19. Artificial insemination allows wider access to highquality male donors.
20. Both of these processes allow for the introduction of
genetic variation from different parts of the world.
21. A vector carries the gene of interest into the
foreign cell.
22. They are much less expensive to produce in
large quantities.
23. A company may use embryo transfer so that they can
choose desirable characteristics for their animals,
and shipping embryos is much easier than shipping
animals. This also allows offspring to grow up in their
permanent environment.
24. Sample answer: The bacteria do or do not do something
to the protein that would normally occur in the
human tissue.
Caption Questions
Figure 4.4 (Student textbook page 162): Unequal
distribution in anaphase could lead to unequal distribution
of chromosomes between new cells. This would lead to
chromosomal disorders or cells that were not viable.
Figure 4.10 (Student textbook page 167): They are
homologous because they are two X chromosomes that are
the same length and carry genes for the same traits at the
same location.
Figure 4.12 (Student textbook page 169):
six chromosomes
Figure 4.13 (Student textbook page 171): Meiosis I would
produce cloned cells, resulting in less genetic diversity
introduced into the cell and the organism.
Figure 4.25 (Student textbook page 184): The vector DNA
contains DNA only from one source, while recombinant
DNA contains the vector DNA and the DNA being cloned.
Figure 4.26 (Student textbook page 185): The nucleus is
removed from the egg cell so that it will only contain the
nucleus with the desired (inserted) genes.
Figure 4.28 (Student textbook page 187): Reduced crop
loss due to disease will result in larger harvests. This saves
the farmer money and allows lower prices.
Figure 4.29 (Student textbook page 188): Students may
suggest that animal genes should not be manipulated for
human benefit, period; that the welfare of the animals
should be considered; or that any method available
should be used to alleviate human suffering. Any answer
supported by an argument is acceptable.
Section 4.1 Review Questions
(Student textbook page 168)
1. All living things are composed of one of more cells.
Cells are the smallest units of living organisms. New
cells come only from pre-existing cells, by cell division.
2. Sample answer: Skin cells undergo mitosis more
frequently than nerve cells do. The skin cells will divide
and repair the cut before the nerve cells are replaced.
3. In a child, because a child is growing in addition to
repairing and replacing tissue
4. G1 is a period of rapid growth and normal cell duties.
S is a period of DNA replication so that each cell
produced receives a full set of DNA. During G2, cells
prepare for division; organelles are made so that new
cells have full complement of information.
5. a. anaphase
b. prophase
c. telophase
6. Sketches should look like Figure 4.4.
7. Sample answer: DNA replicates and I am connected to
my twin; we condense to make chromosomes, line up
along the equator, are pulled apart to opposite sides,
and are then surrounded by a new nuclear membrane.
We relax and get to work.
8. Mitosis duplicates and divides the nuclear material
and cytokinesis divides the cytoplasm, producing two
daughter cells.
9. Daughter cells are genetically identical to parent cells
(except in the case of a mutation).
that when cytokinesis occurred, there would be a
completely random division of the genetic material or
that cell division would stop.
11. The error most likely occurred during anaphase, when
the chromosomes are divided for the new daughter
cells. The centromere did not divide or spindle fibres
only formed from one centrosome, resulting in all of
the chromosomes moving to the same pole. When
cytokinesis occurred, all of the chromosomes were
on one side of the cell. The other daughter cell had
no chromosomes.
12. Students should sketch the basic structure of DNA,
indicating nucleotides and that a certain section of
the DNA contains a gene. Thin threadlike substances
should represent chromatin, and something resembling
the chromosome structure. Sketches should show
that DNA is a component of both chromatin and
chromosomes, such as in Figure 4.7 on student
textbook page 165.
13. Chromosome pairs are not necessarily identical (due
to copying errors, etc.) but the pairs are homologous,
meaning their genes are in corresponding locations but
may have different coding.
14. Diagrams should combine Figures 4.5 (on student
textbook page 163) and 4.9 (on student textbook
page 167).
15. Diagram should show an entire set of chromosomes,
with three that are identical in both size and banding.
16. The X and Y chromosomes determine the sex of the
individual; a female has two Xs and a male has one X
and one Y.
17. a. This is a karyotype, prepared by collecting a cell
sample that is treated to stop cell division during
metaphase of mitosis, then stained to produce a
banding pattern on the chromosomes that is clearly
visible under a microscope. Chromosomes are then
sorted and paired. The autosomes are numbered 1
through 22 and the sex chromosomes are labelled
as X or Y.
b. Male
c. Yes, this individual has the correct number of
chromosomes (23 pairs, 46 altogether).
18. By matching homologous chromosomes (which should
be the same size and pattern), the doctor would see
that one chromosome is shorter than it should be.
10. Since microtubules are responsible for the accurate
division of the chromosomes, it is likely that the
chromosomes would not move to the poles, and
Biology 11 Answer Key Unit 2 • MHR TR
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entanglement of chromatids during prophase 1 that
leads to pieces of chromosome changing places.
Section 4.2 Review Questions
(Student textbook page 181)
Both lead to variation by creating unique combinations
of chromosomes and genes in the sex cells. Sketch
should combine information from Figures 4.18 and 4.19
on student textbook page 175.
1. a. meiosis
b. fertilization
c. mitosis
2. a. 2n = 64
b. n = 32
c. 32
d. 64
12. Sketches should accurately represent the errors in
chromosome structure as shown in Table 4.1 on
student textbook page 177.
3. fertilization
4. Meiosis produces cells that:
• are genetically diverse; and
• contain half the number of chromosomes (are haploid)
5. a. Metaphase II
b. Sketches should show that the sister chromatids
separate and move to opposite poles of the cell.
c. Different colours illustrate that genetic material was
exchanged during crossing over.
d. 2n = 8
6. Meiosis takes place in the reproductive organs (ovaries
and testes).
7. Homologous pairs line up next to each other to make
a tetrad. This leads to chromosomes becoming tangled
(synapse) and DNA trading places. This leads to
increased variation in the daughter cells.
8. Mitosis and meiosis II are very similar in that the
chromosomes line up on the equator and are separated
during anaphase. The significant difference is that
mitosis separates the chromosomes of a diploid cell,
but in meiosis II the cell is haploid.
9. Sample answer:
Spermatogenesis
rQSPEVDFTNBMF
TFYDFMMT TQFSN
rDFMMTQSPEVDFE rQSPEVDFTIBQMPJE
rPDDVSTJO
HBNFUFTGSPN
UIFUFTUFT
EJQMPJEDFMMT
rTUBSUTXJUI
CZNFJPTJT
TQFSNBUPHPOJVN
rFWFODZUPQMBTNJD
EJWJTJPO
Oogenesis
rQSPEVDFTGFNBMF
TFYDFMMT FHHT
rDFMMBOEQPMBS
CPEJFTQSPEVDFE
rVOFWFODZUPQMBTNJD
EJWJTJPO
rTUBSUTXJUI
PPHPOJVN
rPDDVSTJO
UIFPWBSJFT
10. 25 = 32 distinct gametes
11. Independent assortment refers to the fact that the
orientation of each pair of chromosomes along the
equator is independent of the orientation of the
other pairs (ensuring a mix of maternal and paternal
DNA goes to each pole). Crossing over refers to the
4
MHR TR • Biology 11 Answer Key Unit 2
13. Chromosomes do not separate evenly during nondisjunction. During anaphase II, a centrosome may
not divide, leading to sister chromatids both going to
the same pole. During anaphase I, a spindle fibre may
attach to both homologous chromosomes and pull
them to the same pole. This results in one cell having
one too many chromosomes (called trisomy) or one
too few chromosomes (called monosomy).
14. By looking at the homologous pairings in a karyotype,
a clinician can quickly see if one of the chromosomes
is missing its homologue (monosomy) or has two
homologues/three identical chromosomes (trisomy).
15. Invasive methods of prenatal genetic testing pose a
risk to the fetus. If there are no indications of genetic
abnormality, there is no reason to put the fetus at risk
of miscarriage (i.e., invasive methods can be avoided).
16. Students’ arguments should be supported by
statements from either the textbook or individual
research. Students will likely find that it was difficult to
be entirely in favour of or against this type of testing,
illustrating the dilemmas faced by many.
Section 4.3 Review Questions
(Student textbook page 190)
1. Selective breeding. Farmers choose the best animals to
breed with each other by looking at the traits that they
considered favourable (for example, the fastest or those
which produced the most milk).
2. Ways to produce these traits should relate to finding
parent animals with the same traits. Students may
mention selective breeding or artificial insemination.
3. Answers should contain supporting information
such as:
In favour—anything that makes us better should
be encouraged; taking the best characteristics and
allowing people to pick and choose their skills will
make us more productive
Opposing—it is not right to interfere with nature;
everyone could end up being highly skilled in the same
areas and we wouldn’t have people who would want to
or be able to do other jobs
4. Both embryo transfer and IVF involve in vitro
fertilization of egg by sperm. In humans, there is
usually no genetic basis for choosing the egg and
sperm donor. Most human IVF procedures are
undertaken as a result of a fertility problem. Usually in
humans, the embryo is implanted into the female that
donated the egg, whereas in animals, the embryo is
usually implanted in an unrelated female
5. Vectors act as carriers of DNA that a scientist wants to
clone, enabling that DNA to be copied to a foreign cell.
This is important for applications such as gene therapy
and making insulin.
6. Gene cloning copies a segment of DNA, usually for the
purpose of protein production or study. Therapeutic
cloning produces genetically identical cells, usually for
medical treatments. Reproductive cloning produces
genetically identical individuals.
7. Flowcharts and diagrams should accurately represent
the process as shown in Figure 4.25 on student
textbook page 184.
8. Binary fission; some cells underwent conjugation or
had vectors inserted into them
9. Somatic cell nuclear transfer uses an egg cell (with
its nucleus removed) and the nucleus from a somatic
cell. The daughter cells are genetically identical to the
somatic cell.
10. Producing insulin through transgenic plants is
less expensive.
11. The animals produced through reproductive cloning
suffer from health problems and reduced lifespan, and
many are not even born alive.
12. He can choose to breed his schnauzer with another
prize-winning dog (selective breeding) or he could
clone it. Cloning would produce the most exact copy.
13. Stem cells are used in regenerative medicine because
they are undifferentiated. When they are placed into
a patient, they can be stimulated to differentiate and
replace the defective cells of the patient.
14. Students may choose examples such as transgenic
animals like goats which are designed to produce
medical protein products like HGH in their milk,
or pigs that can act as organ donors.
15. a. Sample answer: The Canadian government should
consider the benefits of the transgenic carrots
over regular carrots: are the worms and insects a
significant crop risk? Will it cause more economic
success? Where did the genes come from? Are the
genes naturally-occurring in the area? Will pesticide
eliminate desirable bugs as well? Might the pesticide
genes cross over to other plants? Are there any
possible health risks to those who eat the carrots?
b. Sample answer: The biggest advantage is that
crops will be larger and of a better quality. Other
advantages may be that fewer pesticides are needed,
and that pesticides are delivered at the site of
concern and in the smallest possible amounts,
reducing toxic runoff. Disadvantages might include
a higher seed cost, resulting in higher market prices
and fewer sales. The public may also be reluctant to
buy transgenic food.
c. Answers should show an understanding of the
arguments mentioned in the subsection entitled
“Regulating the Use of Transgenic Organisms” on
student textbook page 188.
16. Students may agree or disagree. If they agree,
arguments might include that it will save lives (prevent
deadly allergic reactions). If they disagree, arguments
might include that there are more pressing issues
than creating a peanut that doesn’t cause an allergic
reaction, or that the causes of allergies are more
complex than a simple allergen–reaction relationship,
and that other solutions are possible.
Chapter 4 Review Questions
(Student textbook pages 195–7)
1. d
2. d
3. e
4. e
5. c
6. a
7. e
8. b
9. growth, repair, and maintenance
10. The bases in DNA connect to each other in a way
that looks like the rungs of a ladder. The sugar and
phosphate make the rails of the ladder, joining the
rungs. The comparison is limited because a real ladder
is flat but DNA is twisted in a helix.
Biology 11 Answer Key Unit 2 • MHR TR
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11. Both have 22 homologous pairs of chromosomes
(called autosomes). The difference is in the sex
chromosome pair: males have one X and one Y
chromosome but females have two X chromosomes.
22. a. Students may name any tissue that undergoes
growth or repair such as root tips in plants or
embryo cells in animals.
b. oogonium or spermatogonium
12. Diagram B represents metaphase I of meiosis because
the homologous pairs are lined up across from each
other on the equator. Diagram A represents metaphase
of mitosis because the homologous pairs are lined up
along the equator.
23. In females the sex chromosomes are a homologous pair
(XX); in males they are not (XY). The X chromosomes
contain the same genes in the same locations. X and Y
do not contain the same genes in the same locations
and are therefore not homologous.
13. haploid cells and cells that are genetically different
24. The micrograph shows crossing over of non-sister
chromatids which is significant because it causes
genetic variation.
14. Oogenesis produces only one egg because of
asymmetrical cytoplasm division, which ensures
enough cytoplasm to nourish the fertilized egg. Sperm
do not need to have cytoplasm for nourishment, so
each of the four cells produced during spermatogenesis
can become a sperm cell.
15. When non-disjunction occurs during anaphase I,
both sets of homologous chromosomes move to the
same side of the cell. If non-disjunction happens in
anaphase II, the sister chromatids do not separate.
Non-disjunction can result in trisomy disorders such
as Down syndrome or monosomy disorders such as
Turners syndrome.
16. Answers should show an understanding of the four
types of errors described in Table 4.1 on student
textbook page 177.
17. If somatic cells can be induced to behave like stem
cells, then embryonic stem cells will not be needed to
continue stem cell research, thus avoiding a significant
ethical dilemma in this type of research.
18. One adult donates an egg, another donates a somatic
cell, and a third acts as the surrogate, carrying the
clone. The clone is genetically identical to the adult
who donates the somatic cell.
19. Applications of transgenic organisms include: drug
production, environmental clean-up, providing organs
for transplant, and improving the food supply.
20. Coiled chromatin is less prone to damage or tangling
while chromosomes move around the cell. During
interphase the uncoiled chromosomes are protected
inside the nucleus.
21. Because A–T and C–G are complimentary pairs, when
the DNA is unzipped for replication, each base acts
as a template, attracting its complement. This results
in two identical DNA strands. If the strands were not
identical, they would not contain the same alleles and
therefore not code for the same proteins; the daughter
cells would not be identical.
6
MHR TR • Biology 11 Answer Key Unit 2
25. Genes would move from one chromosome to another,
which could reduce the cell’s function.
26. a. 24 = 16 distinct gametes
b. 28 = 256 distinct gametes
c. 23 = 8 distinct gametes
27. The major difference between selective breeding
techniques and genetic engineering is that genetic
engineering deals with the genetics of the organism
on a gene-by-gene basis, altering genes if they are
not desirable.
28. Chromosomes that are not homologous may not line
up properly during metaphase, or may not separate
properly during anaphase, which could lead to nonviable gametes.
29. Student answers should include chromosomes, spindle
fibres, and centromeres. They will also likely include
cell membrane, nuclear membrane, and nucleolus.
Props might include balls or balloons, string and pipe
cleaners, or modelling clay, or a floor-sized template for
human “props” to move on.
30.
Mitosis
rQSPEVDFTUXP
EJQMPJEDFMMT
UIBUBSF
HFOFUJDBMMZ
JEFOUJDBMUP
UIFQBSFOUDFMM
rQSPEVDFT
EBVHIUFSDFMMT
rDISPNPTPNFTMJOF
VQBUUIFFRVBUPS
rTFQBSBUFEXIFO
TQJOEMFñCSFT
TIPSUFO QVMMJOH
UIFDISPNPTPNFT
UPUIFQPMFT
Meiosis
rQSPEVDFTGPVS
IBQMPJEDFMMTUIBU
IBWFIBMGUIF%/"
PGUIFQBSFOUDFMM
rHFOFTSFDPNCJOFE
EVSJOHDSPTTPWFS
31. Diagrams should look similar to Figure 4.9 on student
textbook page 167.
32. Diagrams should look similar to Figure 4.20 on student
textbook page 178.
33. Pamphlets should show ultrasound, maternal blood
tests, and amniocentesis such as in Figure 4.22 on
student textbook page 180. Information should include
the value of each type of test, how it is performed,
the possible information gathered from the test, and
the risks.
34. Sample answer:
Method
Selective
breeding
Description
Breeding of
individuals
selected because
of their desirable
traits
Pros/Cons
Imprecise because of the
many genes are involved
and their dominant/
recessive nature
Artificial
Transfer of semen
insemination mechanically/
manually into
a female’s
reproductive tract
Makes semen from highquality males more widely
available; does not require
mating
Embryo
transfer
Transfer of
fertilized egg
manually/
mechanically
into a female’s
reproductive tract
Embryos can be shipped;
offspring raised in natural
environment have more
success
Recombinant DNA
that contains a
specific gene is
inserted into the
host
Useful genes can be
added to other organisms
(e.g., silk-producing
goats can be made by
inserting spider genes into
mammary glands of goats)
Gene
cloning
40. a. to c. A list of genetic tests is available at www.
scienceontario.com. Arguments could include
whether or not the test allows for a change in
behaviour (having offspring) or a change in treatment
(early diagnosis leads to better treatment).
41. Answers should show consideration for the welfare of
the fetus and the ill child.
42. Genetic engineering could allow genes to be inserted
into oranges that would make them grow in Ontario’s
colder climate.
43. Students might research: somatic gene therapy, AIDS
tests, prenatal diagnosis, drugs (e.g., insulin), human
growth hormone, rTPA, hepatitis B vaccine, or
hemophilia clotting factors. Answers should include
the purpose of the drug and its benefits and risks.
A list of potential topics is available at www.
scienceontario.com.
44. Look for evidence that students have integrated their
new learning, synthesizing the information into
their opinions.
45. If stem cells can be used to generate tissues and organs
for transplantation, enough organs could be grown to
meet the demand.
Chapter 4 Self-Assessment Questions
(Student textbook pages 198–9)
1. a
2. c
35. Organizers should include benefits such as
regenerating nerve cells, treating heart disease, regrowing tissue and limbs, testing new drugs, and
growing organs for transplant. Risks of stem cell
research applications might include the ethics of
human cloning.
36. Diagrams should illustrate crossing over and
independent assortment.
3. a
4. c
5. d
6. c
7. e
8. e
9. a
37. Answers should include details about the research (as it
applies to this unit), possible benefits to society and the
environment, and possible consequences.
10. b
38. The graphic organizer should be in the form of an idea
web, a spider map, a fishbone diagram, or a concept
map. The information can be related to the Big Ideas
found in the unit opener or the section titles. Student
answers should reflect the Key Concepts and the
Key Terms.
12. One benefit of cuttings is that they produce
identical offspring. This is also a risk, as all offspring
will be vulnerable to the same diseases and
environmental conditions.
39. Artificial chromosomes would have to be able to
attach to spindle fibres, so would have to have a
functioning centromere.
11. Sketches should be similar to Figure 4.7 on student
textbook page 165.
13. The number of chromosomes is reduced by a factor of
two; 2n becomes n, diploid becomes haploid.
Biology 11 Answer Key Unit 2 • MHR TR
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14. Student diagrams should compare interphase to
metaphase as in Figure 4.4 on student textbook
pages 162–3. They should show that DNA is in the
form of chromatin during the stage of the cell cycle
known as interphase, which includes the G1, S, and
G2 phases. During metaphase, however, DNA is highly
condensed in the form of chromosomes.
15. Sample answer:
chromatid
spindle fibre
centriole
(or centrosome)
centromere
16. Students need not create a graphic organizer; however,
one is provided to summarize all possible differences.
Characteristic
Meiosis
Mitosis
Number of cells
produced
4
2
Number of
chromosomes
n (haploid)
2n (diploid)
Number of divisions 2
1
Variation
All different
Identical
Location
Germ cells
Somatic Cells
Product
Gametes
Somatic Cells
Purpose
Sexual
reproduction
Repair, growth,
replacement,
reproduction
17. Sample answer:
blonde
black
brown
blue
18. A horse will produce gametes that have
32 chromosomes. A donkey will produce gametes with
31 chromosomes. When the two combine in meiosis,
the diploid number is odd; not every chromosome has
8
MHR TR • Biology 11 Answer Key Unit 2
a homologue. This means that meiosis cannot occur, so
no sex cells can be produced.
19. a. Karyotyping uses a microscope to observe cells
during metaphase of mitosis. The chromosomes
were then sorted by length, banding pattern, and
position of the centromere.
b. This is a male with one too many X chromosomes
(XXY). He has Kleinfelter’s syndrome.
c. Non-disjunction during either anaphase of meiosis
causes Kleinfelter’s syndrome. The chromosomes do
not separate properly, resulting in one cell having
too many chromosomes and another having too few.
20. a. Trisomy 21
b. Non-disjunction during either anaphase of meiosis;
the chromosomes do not separate properly, resulting
in one cell having too many and another having
too few.
21. Students may suggest that foreknowledge about
disorders helps parents prepare for the required
lifestyle or that it creates peace of mind if there are
no disorders. Students may also suggest that testing
may result in pregnancy termination, discrimination,
or pose a danger to the fetus (if test is invasive). All
answers are acceptable.
22. IVF → mitosis → one cell analyzed by PGD → only
healthy embryos are implanted
23. Diagrams should be similar to Figure 4.25 on student
textbook page 184.
24. Answers should include some of:
Benefits—increased nutritional value, production of
medicines, increase food supply
Risks—disease transfer, animal rights issues, need to
control growth with stronger chemicals that might
harm the environment, long-term health effects are
not known. Students may suggest limitations such
as the inability to fully test organisms or the need to
label foods to explain that they are transgenic. Any
limitations based on the science are acceptable.
25. In therapeutic cloning, stem cells can be programmed
for many different purposes. For example, to be bone
marrow cells for cancer treatment, nerve cells to
treat neurological diseases, cardiac cells to treat heart
disease, or pancreatic cells to treat diabetes. Ethical
concerns involve the use of embryos as a source of
stem cells, and the question of how long and under
what conditions life should be prolonged.
Chapter 5 Patterns of Inheritance
16. All of the plants are tall with purple flowers.
Learning Check Questions
(Student textbook page 223)
17. See Figure 5.11 (student textbook page 219) for
sample graphics.
a. empty square
b. shaded circle
c. horizontal line connecting male and female
(Student textbook page 205)
1. A cross of two individuals that differ by one trait
2. 3:1
3. The F1 generation exhibits only the dominant form
of the trait, and the F2 generation exhibits both the
dominant and recessive forms in a 3:1 ratio.
4. Mendel needed to start with plants with predictable
traits so that results were accurate and so that he was
able to control his experiments.
5. Student drawings should be similar to Figures 5.3
(student textbook page 204) and 5.4 (student textbook
page 205), but showing wrinkled and round seeds.
6. Brown is the dominant eye colour. Each parent has
the allele for brown eyes. If a child inherits at least one
allele for brown eyes, then the child will have brown
eyes. If a child inherits two recessive alleles, the child
will have blue eyes.
(Student textbook page 212)
7. a. T and t
b. G
c. f
8. 3:1
9. 3 round seeds:1 wrinkled seed
10. Sample answer:
B
b
B
Bb
Bb
b
Bb
bb
18. The gene for the trait is carried on one of the
autosomal chromosomes.
19.
I
II
1
1
2
2
3
20. Two unaffected parents having an affected child
21. Because the lethal disorder is an autosomal recessive
disease, people with only one recessive allele will carry
the gene without getting sick, passing the gene to
their offspring.
22. Since affected parents can have an unaffected
child, woolly hair must be an autosomal dominant
phenotype. Both parents must have one recessive hair
gene, if a child has non-woolly hair. The genotypes of
the children with non-woolly hair are certain, they
must be doubly recessive (ww). You cannot be certain
of the genotype of the child with woolly hair as only
one dominant allele is necessary to produce woolly hair
(W_). Sample pedigree:
I
1
2
11. 210 round-seeded plants:70 wrinkled-seed plants
12. Yes, if both parents are heterozygous for dimples, their
child could be non-dimpled.
(Student textbook page 215)
13. A monohybrid cross involves a cross between two
individuals that differ in one gene (trait). A dihybrid
cross differs by two genes (traits).
14. a. TG, tG, Tg, and tg
b. ABc and Abc
15. Sample answer: Plants and animals are made of many
genes and many traits, and they work together to
provide a healthy animal or plant. Therefore, it makes
sense to study the inheritance pattern of groups
of genes.
II
1
2
3
Caption Questions
Figure 5.3 (Student textbook page 204): The green form
seems to have disappeared. The allele for green is still on
a chromosome, it is just not expressed. The chromosome
doesn’t disappear.
Figure 5.7 (Student textbook page 209): Because purple
is a dominant allele, the purple plant may still carry the
recessive allele. The Punnett square shows the possible
offspring. Just because there are four offspring possibilities
doesn’t mean that all four will occur from a set of parents.
Biology 11 Answer Key Unit 2 • MHR TR
9
Figure 5.8 (Student textbook page 213): Plants of genotype
Y_R_ (yellow round) to Y_rr (yellow wrinkled) to yyR_
(green round) to yyrr (green wrinkled) is 9:3:3:1. The ratio
of YY:Yy:yy is 1:2:1 and the ratio of RR:Rr:rr is 1:2:1.
8. Sample answer:
Parent Phenotypes:
Figure 5.10 (Student textbook page 217): They would
likely be inherited together.
Figure 5.13 (Student textbook page 221): Individual II-2
must be heterozygous because one offspring is unaffected.
Section 5.1 Review Questions
(Student textbook page 207)
1. Dominant refers to an allele that will be expressed
when present alone (e.g., freckles). Recessive refers to
an allele that will only be expressed in individual who
have two alleles for that trait (e.g., no freckles).
2. Traits are determined by pairs of alleles that segregate
during meiosis so that each gamete receives one allele.
Gametes combine to form a zygote that contains alleles
from both the mother and the father. Alleles in the
gamete that were not expressed in the parent are the
recessive alleles. In the offspring, if the zygote receives
two alleles for the recessive trait (e.g., grandmother’s
hair), the trait will be expressed.
3. They support the Mendelian ratio because the ratio
of second-generation plants is 149 tall to 53 short, or
roughly 3:1.
4. a. yellow
b. A heterozygous parent may have been used.
5. Cross a true-breeding purple flowered plant with
a true-breeding white flowered plant. All of the
flowers in the F1 generation will exhibit the dominant
allele’s trait.
6. Phenotype is the term applied to the physical
expression (traits) of alleles and genotype refers to the
combination of all alleles present in an individual.
7. Heterozygous means one of each type of allele for one
or more genes and sufficiently describes the genotype.
To describe the genotype of a homozygous individual,
you need to know whether they have two alleles of
the dominant form, such as YY, or two alleles of the
recessive form, such as yy.
10
MHR TR • Biology 11 Answer Key Unit 2
tall
Parent Genotypes:
homozygous
dominant
(TT)
short
homozygous
recessive
(tt)
F1 Phenotypes:
all tall
F1 Genotypes:
Tt
(heterozygous)
F2 Phenotypes:
_3 are tall
4
F2 Genotypes:
_1 are short
4
Tt (heterozygous), TT (homozygous dominant),
tt (homozygous recessive)
9. a. Sample answer: F = freckles; f = no freckles
b. ff
c. Either FF or Ff is possible because F is dominant to f.
You can only be sure by knowing the phenotypes of
their parents or offspring and looking for a recessive
individual. For example, if one parent has a recessive
trait, the offspring must be Ff.
10. a. Black coat colour is dominant to white coat colour.
b. Each parent must be Bb and the offspring is bb.
11. a. RR
b. rr
c. Rr
12. a. tall
b. tall
c. short
1
2
2
4
c. Genotypes: _
TTRR, _
TtRR, _
TTRr, _
Section 5.2 Review Questions
(Student textbook page 218)
16
16
1. Sample answer:
T
t
T
TT
Tt
t
Tt
tt
BB
Bb
Bb
bb
3. Yes, because the father will pass on the dominant curlyhair gene to some offspring, regardless of whether he is
(Cc or CC).
cc
Cc
cc
16
16
16
10. The ratio is close to 9:3:3:1, so the parents were
likely PpSs.
9
3
11. long and grey: 144 _
; long and black: 48 _
; short
( )
16
3
1
_
; short and black: 16 _
and grey: 48
16
16
( )
c. 25%
Cc
16
16
16
b
c
16
16
B
c
16
3
1
_
short and red, and _
short and yellow
2. a. Both parents are Bb.
B
b
b.
c
16
9
3
tall and red, _
tall and yellow,
Phenotypes: _
A Punnett square distributes the genes from each
parent in all possible combinations.
C
16
1
1
2
2
1
TTrr, _
ttRR, _
ttRr, _
Ttrr, and _
ttrr
TtRr, _
C
C
c
Cc
Cc
c
Cc
Cc
4. A test cross is used to determine the genotype of an
individual with the dominant phenotype.
5. By doing a test cross with a white pig, the breeder will
be able to determine that the pig is heterozygous if any
white piglets are produced. Alternatively, the breeder
could examine the pedigree of the unknown pig.
6. A Punnett square must represent all possible
combinations of the four gametes from each of the
individuals; there are 16 possible combinations.
7. The law of independent assortment states that the two
alleles for one gene assort independently of the alleles
for other genes during the formation of gametes. If
you followed two traits such as plant height and flower
colour, the allele for plant height received by a gamete
during meiosis would not influence which allele for
flower colour was received.
8. a. Tall with purple flowers
b. TtPp
c. TP, tP, Tp, and tp
9. a. All F1 plants will be TtRr; tall with red fruit.
b. TR, Tr, tR, tr
( )
( 16 )
12. Student answers should note the following: genes are
carried on chromosomes. The two different alleles of
any specific gene are carried on the two homologous
chromosomes. When the homologous chromosomes
separate during anaphase I of meiosis, the two alleles of
any gene on that chromosome move into two different
cells. This process corresponds with the concept of
Mendel’s law of segregation.
When all of the pairs of homologous chromosomes
are separating during anaphase I, they do so
independently. That is, different chromosomes have no
influence on each other with respect to which cell the
homologous chromosomes migrate into. Therefore, the
combinations of alleles of different genes that migrate
into one cell or the other is random. This process
corresponds with the concept of Mendel’s law of
independent assortment.
Section 5.3 Review Questions
(Student textbook page 227)
1. mechanism of inheritance of a trait, to determine
the genotype of past generations, and to predict the
genotype of future offspring
2. Autosomal recessive traits can remain hidden in a
carrier. Because they are caused by a recessive allele,
only individuals with two copies will display the trait.
This can only happen 25% of the time when two
carriers produce offspring.
3. autosomal dominant
4. The CFTR gene that causes cystic fibrosis was identified
by Lap-Chee Tsui, a geneticist working at the Hospital
for Sick Children, in Toronto.
5. The genetic test to detect the presence of a CFTR allele
for cystic fibrosis is only 85–90% accurate because
there are so many mutations that cause cystic fibrosis.
Her spouse must have had an allele that was not tested
for or detected.
Biology 11 Answer Key Unit 2 • MHR TR
11
6. Sample answer:
Pros
Cons
• plan ahead for healthcare
• increased guilt
• plan whether or not to
have children
• discrimination in hiring
and for insurance
• prevention or early
treatment
• reduced genetic diversity
7. a. autosomal recessive
b. I-1, II-3, II-4, and III-3 are aa; I-2, II-1, and II-2 are
Aa; III-1, III-2 and III-4 may be Aa or AA because
both parents are heterozygous and the gametes sort
independently. There is a 50% chance that they are
Aa and a 25% chance they are AA. (Students may
note that since we already know from the pedigree
that III-1, III-2, and III-4 are unaffected, there is no
chance of them being aa. Therefore, the probability
1
could be calculated as being _
chance of being AA
3
2
_
and probability of being Aa.)
3
8. The last student is correct; the pedigree could represent
either situation. There is not enough information to
determine whether the pedigree is recessive (two nonaffected parents having an affected child) or dominant.
More generations are required.
9. The parents, I-1 and I-2 are both heterozygous, having
normal pigmentation (Aa). The children who are
albino have the genotype aa. The other two children
could be either AA or Aa. Two-thirds of the non-albino
genotypes are heterozygous.
1
Practice Problems
(Student textbook page 212)
1. a. A only
b. A and a
c. a only
d. A and a
2. Half the offspring will be Pp (purple) and half will be
pp (white).
3. All plants produced will be heterozygous (Gg) with
green pods.
4. Long-winged LL offspring will make up 25%, 50% will
be Ll (long-winged), and 25% will be ll (curly-winged).
5. 50%
7. a. Both parents are Nn.
b. Probabilities are the same: 25% nn and 75% Nn.
2
8. nn and Nn
1
2
3
4
10. Karyotyping is used to identify abnormalities such
as additions, deletions, and translocations, at the
chromosome level. FISH can be used to look for a
smaller abnormality in a chromosome. Genetic testing
is used to identify a specific gene mutation.
11. A genetic counsellor may use a pedigree to determine
the genotypes of family members, and to explain
options for genetic testing, the probability of passing
on a disease-causing allele, symptoms, and available
treatments. They may also provide emotional support
for family members.
12
13. Students may agree, citing that gene therapy could
replace a faulty gene that causes a genetic disease.
Or, students may disagree, citing that there have been
a number of problems (such as immune response
and short-lived treatments) that have not yet been
solved. Either answer is acceptable, when supported
by arguments.
6. The only possible parent combination is Ss and ss.
I
II
12. Gene therapy involves inserting a gene into a host cell
using a vector, as do all gene cloning techniques. Like
other forms of cloning, the hope is that cells that take
up the gene will reproduce and produce a normal cell’s
product. Unlike many forms of cloning, human cells
do not take up the normal gene very effectively, so gene
therapy is not as successful as other types of cloning.
MHR TR • Biology 11 Answer Key Unit 2
9. The parents may be either Bb or bb (but not BB) since
the white pig must have inherited a recessive b gene
from each parent. It is unlikely that the parents are
BB and Bb since a ratio of 3:1 black:white would be
expected if both parents were heterozygous (Bb).
Since black pigs were produced, the parents cannot be
homozygous recessive (bb). To be sure, the phenotype
of the grandparents would need to be known, or if the
pedigree could be re-evaluated after more offspring
are born, to see if the ratio gets closer to 3:1 (implying
parental genotypes of Bb and Bb) or 1:1 (implying
parental genotypes of Bb and bb).
10. GG and gg
(Student textbook page 216)
11. a. Pptt × PpTT, where P = purple, p = white, T = tall,
and t = short
b. parent 1 can produce Pt and pt gametes and parent 2
can produce PT and pT gametes.
12. Black is dominant to brown, and short hair is dominant
to long hair, since brown hair and long hair were not
expressed. The parents were most likely BBss and bbSS.
13. 3/16, or 18.75%
14. a. All of the F1 generation will have short black hair.
b. One in 16
15. a. 50% curly, dark-haired, and 50% straight, darkhaired
b. We can know the genotype of the parents by
examining the phenotypes of their ancestors.
16. GGRr, where G = green, R = round, and r = wrinkled
17. 25%
18. Although both sets of parents would produce the set
of offspring, the ratios are closer for a mating between
one plant heterozygous for both genes and one plant
homozygous recessive for both genes.
19. a. Curly hair must be dominant since the straight hair
trait was not expressed in the parent generation but
reappeared in the children.
b. A cleft chin must be dominant since smooth chin
was not expressed in the parent generation but
reappeared in the children.
c. Both parents are HhCc, where H = curly hair and
C = cleft chin.
1
, or 6.25%
d. _
16
1
, or 25%
20. _
4
Chapter 5 Review Questions
(Student textbook pages 235–7)
1. d
2. c
3. b
4. d
5. a
6. c
7. c
8. d
9. Reasons to use pea plants for breeding trials include:
their rapid growth means many generations are
produced in little time; they have seven distinct
characteristics that are controlled by a single gene
with two alleles each; and it is easy to control
their fertilization.
10. The genotype of the P (parent) generation is
homozygous for the trait, true-breeding, with each
parent displaying the opposite phenotype. Starting with
true-breeding individuals ensures that their genotype
is known.
11. Yes, if the phenotype is the recessive trait, because
an individual displaying the recessive trait must be
homozygous recessive. You cannot tell the genotype
from the dominant phenotype, because the individual
may be either heterozygous or homozygous dominant.
12. A test cross is a cross between a parent unknown
genotype and a homozygous recessive parent. This
is used to determine the genotype of an individual
displaying the dominant trait.
13. 9 (both dominant traits):3 (one dominant trait):
3 (other dominant trait):1 (both recessive traits)
14. a. AB, Ab, aB, and ab
b. ABC, ABc, AbC, and Abc
15. Affected individuals have at least one affected parent;
two unaffected parents only have unaffected offspring.
16. a. rr
b. Rr
c. RR
17. Breed two true-breeding individuals with different
alleles and look for what phenotype their offspring
is. Only the dominant allele will be expressed in the
offspring. The other allele must be recessive.
18. a. black
b. black-haired (BB) and white-haired (bb)
c. black-haired (Bb)
19. a. Dd or DD
b. 50% drooping eyelids (Dd) and 50% non-drooping
eyelids (dd)
c. Probabilities relating to the combination of
alleles apply to each offspring individually, not all
offspring together.
20. a. P male1 = GG, P female = gg, P male2 = Gg
b. 3:1 grey:albino
Biology 11 Answer Key Unit 2 • MHR TR
13
21. One of the parents is heterozygous and the other is
homozygous recessive. The dominant allele could be
determined by breeding each parent with an individual
of the same colour and observing the offspring. For
example, the yellow rat should be bred with another
yellow rat. If any offspring from this cross are black,
you know that yellow is dominant.
22. The diagram displays the results of a dihybrid
cross. The F1 generation display only the dominant
traits (yellow and round). In the F2 generation, all
four genotypes are displayed. This illustrates the
principle of independent assortment; the two genes or
characteristics sort independently during meiosis.
23. The trait is dominant, because two affected parents
have an unaffected child.
24. This shows an autosomal recessive inheritance pattern.
All individuals have a normal phenotype except for I-2,
III-2, and III-3, who are affected (and therefore must
have an ss genotype). Because II-2, II-3, II-4, and II-5
are unaffected offspring of an affected individual, they
must have an Ss genotype. The mother (I-1) may be
either SS or Ss, having passed on the S gene to all her
offspring. II-1 must have the Ss genotype since she is
unaffected yet passed on an s to her offspring. III-2 and
III-3 are affected, so must have the ss genotype. III-1
and III-4 are unaffected, but may be either SS or Ss.
25. Polling must be an autosomal dominant characteristic,
since two individuals with the trait have offspring that
do not have the trait.
L
l
L
LL
Ll
l
Ll
ll
26. a.
I
Test
What the Test
Analyzes
Diagnosed
Disorder(s)
Illustration
Karyotyping
Number of
chromosomes
Turner
syndrome,
Down
syndrome,
Klinefelter’s
syndrome
See Figure 4.21
(student
textbook
page 178)
FISH
Shows
chromosome
abnormalities
CML
n/a
Gene testing
Mutations
Breast cancer n/a
susceptibility
Biochemical
testing
Abnormal
enzymes
Tay-Sachs
disease
n/a
28. Sample answers:
Pro—I believe that genetic testing should be a standard
medical test paid for by OHIP. There is no danger to
the individual and the information gathered from the
test could potentially save the person’s life. Knowing
that you are at risk for cancer means that you can alter
your lifestyle or make sure that you monitor your
health and catch any symptoms early.
Con—I believe that genetic testing should be optional.
Not all people can cope with the stress of a positive
test. The presence of these genes does not mean that
a person will definitely get cancer. If a person wants
to know if they are at risk and are ready to deal with
any result, they should be able to have the test paid for
by OHIP.
29. Students should include gene therapy benefits and risks
discussed in Section 5.3 on student textbook page 226.
1
II
2
1
b. They are both either autosomal dominant or
recessive. There is not enough information to
be sure.
c. The father is DdFf; the mother and son are ddff.
1
d. _
, or 25%
4
14
27. A mind map is well suited to this task. This sample
answer uses a table:
MHR TR • Biology 11 Answer Key Unit 2
30. Students can choose any example but should describe
how meiosis allows for variation (for example, illustrate
the genetic diversity in a litter of kittens).
31. Students may research, for example, sickle cell anemia
or Tay-Sachs disease. Students’ presentations should
describe the disorder and the study, indicate why that
group was chosen, address social and ethical concerns
about this type of research, identify who benefits most
from the research, and make an argument for who
should be allowed access to the genetic information
of the people in the study.
32. The graphic organizer should be in the form of an idea
web, a spider map, a fishbone diagram, or a concept
map. The information can be related to the Big Ideas
found in the unit opener or the section titles. Student
answers should reflect the Key Concepts and the
Key Terms.
33. Does this dog have Von Willebrand’s disease? Do
any of his parents, grandparents, or siblings have
the disease?
34. Probability is not a certainty, especially in the case of
offspring, in which each conception is an individual
event. For each child, the probability that he or she will
display a specific trait is the same (25%, in this case).
The probability that subsequent children will display
a trait is not influenced by the presence of that trait
in previous offspring. The easiest example is boys and
girls. Although the probability of having a boy is 1:2, a
son may be born each time.
35. Students may argue that parents do or do not have a
responsibility to inform children of recessive disorders
they may have inherited. Either answer is acceptable.
Sample answer: This information can help children
decide whether or not to have their own children. If
their spouse has the same gene, being a carrier means
that their children could show the genetic disorder.
For minor disorders such as colour blindness, this may
seem trivial.
36. A list of genetic tests is available at
www.scienceontario.com.
37. a.
I
II
1
2
Chapter 5 Self-Assessment Questions
(Student textbook pages 238–9)
1. a
2. c
3. c
4. c
5. c
6. a
7. a
8. b
9. b
10. d
11. a. Homozygous means having two of the same allele
for a gene and heterozygous means having two
different alleles for the same gene.
b. An allele that is dominant is expressed when there
are one to two copies but a recessive allele requires
two alleles in order to be expressed.
12. The farmer should breed heterozygous cattle with
homozygous dominant cattle. This will produce 50%
heterozygous, with the best meat, and the other 50%
homozygous dominant, which although will not have
the best meat, will not die.
13. 50%
14. Cross the unknown plant with a short plant producing
white flowers (a homozygous recessive plant). If
the offspring include any plants with the recessive
phenotype, then the parent plant must be heterozygous
for that characteristic.
15. Both are heterozygous RrSs.
1
2 (Taku)
3 (Sara)
b. Is there a history of cystic fibrosis (CF) in Sara’s
family? If not, their children will not have cystic
fibrosis since children will not receive the two copies
of the defective allele necessary to produce this
autosomal recessive disorder.
c. Genetic testing to identify whether or not they are
carriers of cystic fibrosis
d. The test is only 85-90% accurate and they may
be concerned by possible use of the test results in
the future.
16. Drawings should show chromosomes sorting during
meiosis, illustrating that genes sort independently
unless they are on the same chromosome.
17. Since recessive traits are only expressed in
homozygotes, the trait may not appear in all
generations. To identify the presence of an allele in a
family, several generations of data must be obtained.
18.
I
II
1
1
2
2
3
Biology 11 Answer Key Unit 2 • MHR TR
15
19. There isn’t enough information to determine the
mode of inheritance. If two unaffected parents had
an affected child, we could tell that the mode of
inheritance is autosomal recessive. If two affected
parents have an unaffected child, we could tell that the
mode of inheritance is autosomal dominant. We need
more information from other generations to determine
the mode of inheritance.
20. Pedigrees should show both parents affected. Children
may or may not be affected despite the 75 percent
probability of them having a peaked hairline.
Sample answer:
I
II
1
1
2
2
3
4
21. a. Cystic fibrosis is an autosomal recessive disorder
caused by one of over 1600 different mutations
to the CFTR gene on chromosome 7. The genetic
tests for this disorder each look for only one
specific mutation.
b. A child inherits the gene from parents, and must
inherit the gene from both parents to be affected.
c. A person with cystic fibrosis has a life expectancy in
the late 40s. If they have the disorder or are carriers
of the disorder, they may not choose to reproduce or
may experience difficulty finding life partners. The
disorder does not have delayed symptoms so there
is no risk of future discrimination because they may
get sick.
22. Any personal opinion is acceptable if it shows evidence
of incorporating knowledge of genetic diseases, testing,
and inheritance.
23. The allele for Huntington disease does lead to a
drastically shortened life span, so it is sometimes
considered to be a lethal disorder. On the other hand,
it may be argued that it is not a lethal disorder since
sufferers normally survive long enough to reproduce.
24. Students should discuss skills that include counselling
and remaining impartial. Counsellors must be able to
explain the alternatives and provide support regardless
of the decisions made by a family.
25. Students should discuss that if perfected, gene therapy
will be able to replace abnormal genes with normal
ones, but that the process is not yet perfected. Problems
with the viral vector have led to some deaths, and even
if not fatal, there has been no long-term success in
human trials.
16
MHR TR • Biology 11 Answer Key Unit 2
Chapter 6 Complex Patterns
of Inheritance
Learning Check Questions
(Student textbook page 244)
1. Incomplete dominance is a condition in which neither
allele completely conceals the presence of the other,
resulting in an intermediate phenotype. Codominance
is a condition in which both alleles in a heterozygote
are fully expressed.
2. Multi-letter notations represent the combination of
allele dominance. Lower case letters are only used to
represent alleles that are recessive to another allele.
Capital letters show that neither trait is recessive to
the other.
3. a. Light purple
b. White and purple
4. Having some cells that are sickle-shaped provides the
carrier with immunity to malaria, a deadly disease
prevalent in some parts of Africa.
5. Sample answer: Red and white flowers that produce
pink flowers suggest an intermediate phenotype, and
codominance is found in roan cows.
6. When scientists were able to take a closer look at the
blood cells in heterozygous individuals, they saw that
both normal and sickle cell blood cells were present,
supporting the idea of codominance.
(Student textbook page 253)
7. Linked genes do not assort independently.
8. Linked genes are found on the same chromosome and
are inherited together; phenotypic ratios that they
produce do not match the Mendelian ratio.
9. The more frequently linked genes get separated,
the farther apart they are on a chromosome. This
provides an idea of the relative positions of genes on a
chromosome, which can be visualized in a gene map.
10. Yes, otherwise the ratio would be 9:3:3:1, which
is expected for unlinked traits resulting from a
dihybrid cross.
11. Sample answer: The genes for the traits are likely to be
located on the X and Y sex chromosomes.
12. Because linked genes can be separated from each
other during meiosis, the gene that is being tested for
may not always indicate the presence of the diseasecausing allele.
(Student textbook page 262)
13. Determined the DNA sequence of the human genome;
identified all of the genes; made the genes accessible to
all, address social, ethical, and legal issues that arise;
and sequenced other organism’s genomes
14. about 2%
15. biological molecule sequencing technology, computer
software, and communication technology
16. Bioinformatics allowed for organization and analysis
of large amounts of sequence data that was generated
from labs all over the world.
17. The Internet allowed for rapid and easy transfer of data
(and communication) between labs all over the world.
18. Sample answer: A new DNA sequence of the genome
for an organism has been identified and they want to
compare that sequence to that of the human genome.
Because this disorder shows intermediate levels of LDL
in heterozygotes, it is incomplete dominance. If it was a
case of codominance, you might see that levels are high
one day and low the next.
3. a. Students should use an uppercase letter for the gene,
for example C, and then use two uppercase letters
for the superscripts, for example C R (red) and
C W (white). The genotypes for the three radishes
would then be C RC R (red), C WC W (white), and
C RC W (purple).
b. 1:2:1 red:purple:white
4. a. Feather colour must be a case of codominance since
each allele is expressed independent of the other.
b. black (C BC B) and speckled black and white (C BC W)
Caption Questions
5. a. Students should draw a blue organism (Bb).
b. Students should draw a green organism (C BC Y).
c. The student should draw an organism, parts of
which are yellow and parts of which are blue (C BC Y).
Figure 6.6 (Student textbook page 245): Agouti: CC, Cc ch,
Cc h, Cc; chinchilla: c chc ch, c chc h, c chc; Himalayan: c hc h, c hc;
albino: cc
6. a. I-1, I-2, II-2, II-3, and II-4-Hb AHb S, II-1 and III-1
are Hb SHb S. All genotypes are determinable.
b. 25%
Figure 6.9 (Student textbook page 248): As the number
of gene pairs involved increases, so does the range of
possible phenotypes.
7. a. Both parents are heterozygoous c chc and c hc.
b. 2:1 chinchilla:Himalayan
Figure 6.10 (Student textbook page 251): PPLL, PpLL,
PPLl, ppll
Figure 6.11 (Student textbook page 252): Alleles that are
close together will likely be transferred on the same bit
during crossing over. The farther apart two alleles are, the
more likely that they will get split up during crossing over.
Figure 6.13 (Student textbook page 254): 50% of females
will be X RX r (red-eyed); 50% of females will be X rX r
(white-eyed); 50% of males will be X RY (red-eyed); 50% of
males will be X rY (white-eyed)
Section 6.1 Review Questions
(Student textbook page 250)
1. a. incomplete dominance
b. codominance
2. Hypercholesteremia leads to a build up of LDL (bad)
cholesterol in the blood. Homozygotes for the normal
allele have the lowest LDL levels, heterozygotes
have an intermediate level, and homozygotes for the
disease allele have the highest levels, demonstrating
the intermediate phenotype for the heterozygote.
In incomplete dominance there is an intermediate
phenotype that is a blend of the extreme phenotypes.
8. 0%; To get an agouti rabbit, you must have at least one
agouti parent since it is the most dominant allele.
9. Sample answer:
ABO Blood Groups
Example
Multiple alleles
I A, I B, and i
Codominance
I A and I B alleles, both are expressed,
producing blood type AB
Dominant/recessive
ii results in O blood type
I Ai and I Bi result in A and B blood types
10. The gene that controls coat colour in Siamese cats
may be influenced by environmental factors including
temperature. Areas of colder temperature (such
as extremities: ears and tail) may lead to a darker
coat colour.
11. The continuous variation expressed in human skin
colour (from very light to very dark) suggests that skin
colour is a polygenic trait.
Biology 11 Answer Key Unit 2 • MHR TR
17
Section 6.2 Review Questions
(Student textbook page 259)
1. Sample answer: Start with true-breeding plants for
both traits. Cross to obtain the F1 generation which
should all display the dominant traits. Cross two plants
from the F1 generation. If the two genes are not linked,
the ratio of phenotypes in the F2 will be 9:3:3:1. If the
genes are linked, the ratio will be closer to 3:1.
2. Genes that are on the same chromosome are linked.
When crossing over occurs between non-sister
chromatids, the alleles of the linked genes can
segregate, or become “unlinked.” Alleles that were once
on the same chromosome are then located on two
different chromosomes.
3. Linked genes would not produce the expected
Mendelian ratio for a dihybrid cross (9:3:3:1).
4. Alleles P and Q are closest on the chromosome, since
the crossover frequency is the lowest.
5. Diagrams should look similar to Figure 6.11 (student
textbook page 252).
6. They reproduce rapidly, produce many offspring, and
sex determination is controlled by XY chromosomes.
7. a. The woman must be a carrier, X CX c, and the man
must be X CY.
b. The colour vision deficient child must be a boy, since
a girl would inherit a normal allele from her father
and cannot have CVD.
8. X hY (father) and X ?X h_ (mother) The father must have
hemophilia since the woman inherited one of her X h
from him and he only has one to give. The mother
must have an X h also, but her other allele could be
X H or X h.
9. A girl with Turner syndrome has only one
X chromosome, so if she inherits an allele for CVD
from her mother and no X chromosome from her
father, she could have CVD.
10. Students will likely disagree. Although there are more
males affected than females, the pedigree does not
show the standard pattern of affected male through
daughter to her son (Individual II-1/II-2). Students
could argue X-linked recessive if they say that II-2 is
a carrier.
11. An autosomal trait will have a comparable number
of males and females affected. For X-linked recessive
traits, there should be more males affected. Both will
show traits skipping generations, but in an X-linked
recessive inheritance pattern, it will skip generations
(from an affected grandfather through the mother to
an affected son).
18
MHR TR • Biology 11 Answer Key Unit 2
12.
I
1
II
1
2
2
3
3
The chance that his sisters are carriers is 50%, since
they inherit a normal allele from their father, and have
a 50% chance of inheriting the disease allele from
their mother.
13. X-inactivation occurs in females, so a female with one
allele for the disease trait will likely have only about
50% of her cells affected.
14. See Figure 6.14 (student textbook page 255) for the
format of the diagram.
15. Because different X chromosomes are deactivated in
patches of cells, some sweat glands will not contain the
disorder and will therefore produce sweat. Other sweat
glands will have an active, recessive X so they will not
be able to produce sweat.
Section 6.3 Review Questions
(Student textbook page 267)
1. Sequencing an organism’s genome makes it
useful in genetic research. Also, knowing other
sequences provides a basis for comparison with the
human genome.
2. Sample answer:
Book
Pages
Genome
Genome
Paragraphs
Chromosomes
Sentences
Genes
Words
Codons
Letters
Nucleotides
3. Facts learned through the Human Genome Project
include: only 2 percent of nucleotides code for genes;
there are 25 000 genes; 50 percent of our DNA is made
of repeating sequences; and about 99.9% of the DNA
sequence is almost exactly the same in all people.
4. Ongoing research includes determining the function
of genes that were discovered and determining the
function of the non-coding repeating regions of DNA.
5. Sample answer: Yes, if it was not for the Human
Genome Project, we would not know the location
of disorders that can be identified through genetic
tests and we would not be able to treat disorders
using gene therapy, or to create insulin through
transgenic organisms.
6. The difference is that with the advent of bioinformatics,
much of the genetic research that takes place involves
the use of computers “(“dry lab”) rather than just “wet
lab” equipment.
7. Sample answer: The components of the term
bioinformatics are “bio” in reference to biology,
“info” in reference to information, and “matics”
in reference to mathematics. Alternative names
include genometrics, biomancy, macrogenetistry,
computational biology, and bionomics.
8. Genomics is the study of genomes and how genes
work together to control phenotype. Multiple genes
are considered at one time and the entire organism is
studied. This compares to Mendel looking at one trait/
gene at a time. This will be helpful in understanding
the inheritance of diseases like cancer and to help
develop treatments for these diseases.
9. The International HapMap Project is an international
sequencing project that is attempting to map variations
in the human genome. The main goal is to provide
researchers with enough information to allow them to
connect specific variations to specific genetic diseases.
10. Epigenetics suggests that gene function may change
without the actual DNA sequence of the gene
changing. In other words, it looks at the mechanisms
that control whether our genes are on or off.
11. By looking at thousands of genes at one time,
scientists can see which genes are active under certain
conditions. This helps them to understand how gene
expression is coordinated.
12. Sample answer:
Benefits
Risks
Medical treatments can be
designed to match a specific
individual’s genome
Can be used to discriminate
against an individual in the
case of insurance
13. Sample answer: No, if the profile is not kept private,
the person could be discriminated against by insurance
companies and employers. Just having a particular
gene does not guarantee that you will get a particular
disorder or disease and the knowledge could cause
undue stress or discrimination.
14. a. Some issues that might be considered are: Who
owns the genetic information? Should companies
have the right to sell DNA information to other
companies without the permission of the people
who provided the samples? Should companies that
use DNA in medical research be required to share
the results of their work with the individuals or
communities whose genetic information was used?
Where is the boundary between public and private
genetic property rights? Should the legal system
have the right to the information?
b. Sample answer: Providing genetic information
should be voluntary. Genetic information can
provide medical benefits to many. Research should
be funded so as to maximize benefit, for as many as
possible. It is reasonable for companies to expect a
return on their investment in genetic research. The
outcome of the research should be widely available.
c. Sample answer: To keep research going, information
could be gathered but kept private through coding
that only the individual will recognize. This will
prevent discrimination while still providing
information to researchers and the individual.
15. Sample answer: I think that people should be
responsible for themselves. If they want to work in
an environment that may potentially prove harmful
to them due to their genetic profile, that should be
the individual’s decision to make. To expect that an
employer use information in the profile to create an
idea work environment yet not discriminate against an
employee for a potential genetic issue is unfair to the
employer and has too great a potential to be misused.
Practice Problems
(Student textbook page 247)
1. A, AB, or B (if the woman is heterozygous)
2. AB or A
3. The baby cannot be theirs because each of her type O
blood parents gave her an i allele. This woman’s baby
would have received either an I A or I B allele from her.
4. Yes, parents with the genotypes I Ai and I Bi could
produce offspring with any of these blood types.
5. c chc and c hc
6. 1:1 chinchilla:Himalayan
7. A mating between a chinchilla rabbit with a c chc h
genotype, and an albino rabbit (cc) has a 50 percent
chance of producing a Himalayan rabbit (c hc).
Biology 11 Answer Key Unit 2 • MHR TR
19
8. a. Sample answer:
Mother
IB
IB
father i
I Bi
I Bi
i
I Bi
I Bi
Mother
IB
IB
father I B
I BI B
I BI B
IB
I BI B
I BI B
b. Students should choose whichever cross produces
a higher percentage of offspring with the genotype
I B_, those with blood type B. In this sample, that
would be either of the sets of parents. An alternative
cross could have been two heterozygous parents
producing 50% of offspring with blood type B.
Students would then choose one of the samples
shown, both of which have 100% of offspring with
blood type B.
9. a. Parent I-1 is A Sa t and parent I-2 is a ya t.
b. 50%
c. A Sa y or A Sa t
17. The inheritance is most likely X-linked dominant since
the daughters of affected males are affected, as are some
daughters and sons of affected females. Affected males
did not pass the trait to their sons.
18. a. All of the female offspring will be normal, the male
offspring will all be hearing impaired. Since this
is carried on the Y chromosome, the information
about the female dog is irrelevant.
b. No offspring will be hearing impaired.
19. Cross the red-eyed female with the male to obtain
carrier females. Cross the F1 female (carrier) with
the original white-eyed male to get half the female
offspring with white eyes.
20. His genotype is LlX cY. Her genotype could be any that
has X C and l. (For example, X CX CLl.) Because none
of their children have CVD, she is not likely a carrier
of CVD but best guess is that she is homozygous
dominant since there are no offspring with long fingers.
10. Sample answer:
dark coloured dog
AS
at
sandy coloured a y
A Sa y
a ta y
t
S t
t t
dog a
Aa
aa
(Student textbook page 258)
11. a. The mother is X CX c and the father is X cY.
b. Out of four possible, one would be a carrier female
(X CX c), one a colourblind female (X cX c), one a
normal male (X CY), and one a colourblind male
(X cY).
Mother
12.
Xc
Xc
c c
c
c c
XX
XX
Y
X cY
X cY
Father X
13. 100%
14. a. X NY (father) and X NX n (mother)
b. The boy is X nY because he is affected by the disease.
The girl must be X NX n or X NX N because she could
get either the dominant or the recessive gene from
her mother and had to get the dominant gene from
her father.
Quirks and Quarks Feature Questions
(Student textbook page 249)
1. autosomal recessive; for an individual to be deaf due
to the mutation they must have two copies of the
mutated gene
2. heterozygous advantage refers to an advantage
that a heterozygous individual has compared to
individuals who are homozygous dominant or
homozygous recessive for the trait. People with one
copy of the mutated gene are not deaf and also gain
an advantage—a thicker layer of skin that could offer
better protection and defence against infection.
3. Sample answer: Knowledge in the following fields
would be an advantage: molecular biology (study
of biology at the level of the molecules in the cell);
genetics; human diseases that are caused by genetic
mutations; and research skills for how to study cells
and the genetic material in cells.
Chapter 6 Review Questions
(Student textbook pages 273–5)
1. b
2. d
15. 50%
3. c
16. a. The pattern of inheritance is likely X-linked
recessive, since the yellow disappeared in the F1, and
reappeared only in F2 generation males.
b. The females are X TX T and X TX t, and the males are
X TY and X tY.
c. 50%
4. d
20
MHR TR • Biology 11 Answer Key Unit 2
5. a
6. b
7. c
8. c
9. a. incomplete dominance
b. codominance
c. red as dominant
10. 1:2:1
19. No, you could not establish true-breeding platinum
foxes, as the genotype for the form is heterozygous.
A cross between C PC p and C PC p (platinum-coloured
fur) parents will produce 25% C PC P (lethal), 50% C PC p
(platinum-coloured fur), and 25% C pC p (silvercoloured fur).
11. Heterozygous advantage describes a situation in
which an individual who is heterozygous for a trait
has a reproductive advantage over individuals who
are homozygous for the trait. An example is sickle
cell anemia in areas of the world that have malaria.
Homozygous dominants are susceptible to malaria,
and homozygous recessives have sickle cell disease,
which reduces their oxygen–carrying capacity. Being
heterozygous with the sickle cell trait increases
resistance to malaria but does not greatly compromise
oxygen capacity as only some of their cells are sickleshaped.
22. The rabbit could have a gene that is activated and turns
its hair black in response to cold.
12. Sample answer: Sometimes there are variations in the
expression of a gene. This can be shown through blood
type inheritance where there is an allele that makes the
A protein on blood cells, a different allele that makes a
B protein and a third allele that makes no protein. Due
to the multiple alleles, there are more than two possible
phenotypes for the trait.
25. You may conclude that the gene for wing shape is
Y-linked. Since the X chromosome is inherited from
the female parent, some of the males would have
normal wings if the gene for wing shape were X-linked.
13. The continuous variation in height suggests that it is
a polygenic trait.
14. Linked genes are genes located close together on a
chromosome. They are not usually separated during
crossing over and therefore are always sorted together
(not independently) during meiosis.
15. Duchenne muscular dystrophy is linked to the
X chromosome, which a boy receives only from his
mother. Any normal mother of affected offspring must
be a carrier.
16. A person’s genetic profile is that person’s complete
genotype (including mutations). It can help doctors
make generalized risk assessments about disease but
could also be used as a tool for discrimination if it is
accessed by insurance companies or employers (for
example).
17. This is likely incomplete dominance since one of their
children has an intermediate phenotype.
18. a. Individuals I-4 and I-6 both have blood type A.
b. No. Since female III-2 has blood type O, her
genotype is ii. The father with AB blood has the
genotype I AI B. Their children can only have
either type A blood (genotype I Ai) or type B
(genotype I Bi).
20. Blood types among the children could be A, B, and
AB. If any child is blood type A then the man must be
heterozygous type B.
21. I AI B, I Ai, I AI A, I BI B, or I Bi
23. The distance between the genes may mean that
crossovers occur often, resulting in frequent
recombination between these genes.
24. None, because female offspring have XX sex
chromosomes and therefore cannot pass a Y
chromosome to their sons.
26. a. Mendel would have seen combinations of traits
not present in the parents and different phenotypic
ratios such as 0.42:0.41:0.09:0.8.
b. Sample answer: Mendel might have hypothesized
that if the traits were carried on the same
chromosome, then a cross between two truebreeding plants for dominant and recessive traits
would produce offspring that only display the
dominant phenotype.
c. Mendel could have crossed the two parental plants,
e.g., maternal pure breeding dominant for both
traits, and paternal pure breeding recessive for both
traits. He may have observed recombinants in the F1
phenotypes, indicating that there was crossing over
of the maternal and paternal genes.
27. a.
I
II
1
2
1
3
4
2
b. There is no chance of having a girl with CVD.
A daughter would either be a carrier, or not have
the gene at all.
c. There is a 50% probability of their son having
normal vision.
Biology 11 Answer Key Unit 2 • MHR TR
21
28. Students should draw or construct a homologous pair
of chromosomes, containing a number of genes. They
should illustrate crossing over, and show that genes
that are close together on a chromosome tend to be
transferred together during crossing over.
Description
Benefit
Concern
Proteomics
and
Topic
Studying
threedimensional
shapes of
proteins and
determining
their functions
Identifying
proteins can
lead to the
identification
of genes
responsible
for certain
diseases
Very complex
to get the DNA
sequence,
as many
variations in
the order of
nucleotides
can result
in the
production
of the same
protein
Epigenetics
Studies how
changes in the
inheritance of
certain traits
or phenotypes
are based
on gene
functions
rather
than gene
sequences
Identifying
the causal
factors
(triggers)
of genetic
diseases that
are latent
29. See Figure 6.11 (student textbook page 252). The left
side of that figure shows the contradiction. The right
side shows independent assortment.
30. Answer could show consequences of discrimination or
positive results of research curing a disease.
31. Sample answer: Nucleotides are like notes, DNA is like
a chord, genes are like phrases, and chromosomes are
like the song.
32. A mind map or wheel organizer is well-suited to this
task. Uses of bioinformatics should be at the centre.
The “spokes” should detail: the analyzing of genomic
data, identifying genes associated with human health,
preventative treatments, and cures for diseases.
33. Sample answer: Read the first column from top to
bottom to see connections between the topics, then
grow out along each row to build concept map.
Topic
Description
Benefit
Concern
Determined
the sequence
of DNA in the
entire human
genome
Showed
how many
genes we
have, how
similar we
are to other
organisms
and each
other
What the
information
could lead to
Bioinformatics Applies
which leads to computer
technology
to create and
maintain
databases of
information
(chemist
Margaret
Dayhoff )
Allows
everyone
easy
access to
information
Genetic
profiles could
be improperly
used
Genomics and
Identifying
complex
diseases
(such as
cancer)
Human
Genome
Project
is a result of
22
The study of
genomes and
the complex
interactions
of genes
that result in
phenotypes
MHR TR • Biology 11 Answer Key Unit 2
Many diseases
involve a
complex array
of factors.
Results could
be misleading
34. Organizers should include all of the Key Terms and
Key Concepts listed in the chapter summary.
35. a. The farmer could selectively breed tall corn plants
and harvest the seeds from the offspring and then
repeat the process again and again.
b. The farmer’s work will be most effective if the height
of corn plants were determined by multiple alleles
or codominance, because then only the tall plants
could be selected to breed instead of working with
varying heights produced by polygenic inheritance.
c. The farmer could cross various parent plants to see if
the offspring of particular crosses show evidence of
the disease resistance.
d. The farmer would avoid selecting any plant for
breeding that showed the disease and then select
only the tall plants for breeding from the diseasefree offspring.
36. a. to c. Answers should show evidence of
independent research.
37. a. to c. Sample answer: The goal of Enhancing Canola
Through Genomics was to improve canola seed and
apply that knowledge to other related crops. It is
important to make sure that the research done does
not have huge consequences.
18. The gene for it occurs on the X chromosome. Males
only have one X and therefore if they inherit the
recessive allele it will always be expressed. For a
pedigree, see the diagram in question 17 on student
textbook page 277.
Chapter 6 Self-Assessment Questions
(Student textbook pages 276–7)
1. d
2. b
3. e
19. The gene for it occurs on the X chromosome. Males
only have one X and therefore if they inherit the
recessive allele it will always be expressed. A carrier
must be heterozygous.
4. b
5. d
6. b
20. Barr bodies are inactive X chromosomes. Fur colour
is X-linked. Because only one X is active at a time, a
heterozygous cat will have some patches of fur with
one X active (and a fur colour associated with that X)
and other patches with the other X active, causing a
different colour.
7. c
8. b
9. c
10. e
11. 1:2:1 of red:purple:white
12. a. All offspring will be green
b. Offspring will be green and yellow; Could be in
patches, could be that some seeds are green and
others are yellow
c. All offspring will be intermediate in colour between
green and yellow––yellowish green or greenish
yellow.
13.
B
Alleles
O
A
A
A
A
O
B
B
B
O
Type A
O
O
21. So that we can compare the human genome to them
and see which sequences are shared, and therefore
which proteins are shared
22. Sample answer: The individual nucleotides are like the
letters in the language. Groups of nucleotides make
words and groups of words make sentences that have
a purpose (this is what a gene is). A chromosome
is composed of many genes, just as a paragraph is
composed of many sentences. If you don’t understand
the structure of the language, you can not decode it.
AC TG AATCG G T TAAC TATCG T TACG
Type O
word/base pair
sentence/gene
Type B
paragraph/chromosome
14. The production of chlorophyll is controlled by
the light turning the genes on or off. Chlorophyll
is essential for the process of photosynthesis. By
researching this, scientists can better understand how
to control the production of chlorophyll and therefore
increase productivity.
23. Genomics research focuses on the interaction of genes
in an organism. By identifying disease- or disordercausing genes, we can look at methods that could cure
the disorders or screen for them.
15. See Figure 6.11 on student textbook page 252. A and
b are linked. They can become unlinked if crossing
over occurs.
25. Variation in non-coding regions can be used in DNA
fingerprints to help identify forensic evidence.
16. 50% red-eyed fruit flies X RX r and 50% red-eyed male
fruit flies X RY.
Unit 2 Review Questions
(Student textbook pages 281–5)
17. a. It is sex-linked because only males are affected. It is
X-linked recessive because two unaffected parents
have an affected child.
b. X HX h
c. 25%
24. Bioinformatics is the production of a research database.
The Human Genome Project uses bioinformatics.
1. d
2. c
3. c
4. c
5. a
6. a
Biology 11 Answer Key Unit 2 • MHR TR
23
7. b
8. e
9. c
10. e
11. G1: rapid growth and cell activity; S: DNA synthesis
and replication; G2: cell prepares for division
12. A karyotype is a picture of the set of chromosomes
that an individual has. It is used to determine sex,
diagnose monosomy and trisomy disorders or
chromosome abnormalities.
13. same size, position of centromere, and banding
patterns; they are not identical because each
chromosome could carry different alleles for the
same genes
14. Haploid cells contain half the set of chromosomes.
They are also called gametes. In humans they are found
in the testes and ovaries. Diploid cells contain a full set
of chromosomes, and are called somatic cells. They are
found in all tissues except the germ tissues.
15. One essential outcome of meiosis is variation in
the cells produced. This occurs during prophase
when crossing over happens and in anaphase when
chromosomes sort to either pole. The other essential
outcome is that the cells produced are haploid.
This occurs during anaphase I when half of the
chromosomes go to one pole and the other half go to
the opposite pole.
16. a. 2 pairs
b. 2 chromosomes
c. 4
17. Pea plants are self-fertilizing (they have male and
female organs), are true-breeding, grow quickly, and
have only two alleles for each trait studied. Each allele
shows simple dominant/recessiveness.
18. Dominant and recessive refer to alleles. Dominant
alleles make proteins and are always expressed in
phenotype. Recessive alleles do not make proteins
and are only expressed when the individual has the
homozygous recessive genotype.
19. Monohybrid crosses examine the inheritance of one
trait. Dihybrid crosses involve two different traits.
Punnett squares show the fertilization possibilities for
each cross. The male gametes are placed along the top
and the female along the side. The intersections have
genotypes entered which represent zygotes.
24
MHR TR • Biology 11 Answer Key Unit 2
20. Autosomal recessive inheritance refers to the
inheritance of a recessive trait that is found on one of
the autosomes (non-sex chromosomes). Cystic fibrosis
is an autosomal recessive disorder. A person will only
have CF if they are homozygous recessive for the trait.
21. Genes are located on chromosomes, which provide the
basis for the segregation and independent assortment
of alleles. Walter Sutton studied the process of synapsis
and migration of chromosomes during meiosis, and
related his observations to the behavior of the factors
described by Mendel.
22. Students should be able to describe any three of the
genetic tests summarized in Table 5.4 on student
textbook page 224.
23. Because there are three phenotypes (one for
each genotype) and the intermediate phenotype
(heterozygous) has some cells that are normal and
some that are sickle shaped
24. When a single gene has multiple alleles, there are more
than two alleles for the trait. Sometimes these alleles
have a dominance continuum and sometimes there is
codominance between some of the alleles. Blood type
is an example. When both dominant alleles are present
in the genotype, the blood is type AB, when just one
dominant allele is present, the blood type is either A or
B, and when the genotype is homozygous recessive, the
blood type is O.
25. Because males have one X chromosome, if they have
the recessive allele it will always be expressed.
26. By collating the data in a central bank, information can
be used by several scientists at one time.
27. Mitosis and meiosis both involve cytokinesis and
interphase (growth and DNA replication) and have
phases called prophase, metaphase, anaphase, and
telophase. Mitosis occurs in somatic cells; meiosis
occurs in germ cells. Mitosis makes two identical cells
that are diploid. Meiosis makes four haploid cells that
are different.
28. A gene is a segment on a chromosome that codes for a
trait. An allele is the form of the gene (each gene will
have different ways of being expressed).
29. a. Mutations (see Table 4.1 on student textbook
page 177) and non-disjunction (see Figure 4.20
on student textbook page 178).
b. Karyotyping and biochemical analysis
30. They don’t; the only difference is that fertilization
occurs internally in artificial insemination and
externally in embryo transfer.
31. See the answer to question 9 in the Section 4.2 Review.
32. a. BB
b. bb
c. Bb
d. B; b; B, b
33. a. Round is dominant since all F1 offspring are round.
b. The data is very close to the Mendelian ratio. I would
expect 75 percent of the offspring to be round and
75.8 percent were round. The percent may not be the
same due to the small number of offspring observed.
34. Yes, if both parents are heterozygous for hairline
35.
Gamete from Gamete from
Genotype
Male Parent Female Parent of Offspring
TY
tY
TtYY
Phenotype of
Offspring
tall plant,
yellow seed
colour
Gt
gt
Ggtt
green pod,
short plant
Yg
yg
Yygg
yellow seed
and pod colour
36. a. Both parents are PpTt.
b. PT, Pt, pT, pt
c. 17
37. a. Dominant. If shaded was recessive, all offspring
would have to be recessive.
b. If A = affected, I-1, I-2 are Aa, II-1 is AA or Aa
(since both parents are heterozygous and the
disorder-causing allele is dominant), and II-2 and
II-3 are aa.
38. a. C RC W (pink); each parent gives one allele. The red
parent can only give red alleles, the white only white
alleles. This makes the offspring heterozygous and
flower colour is incompletely dominant.
b. pink C RC W, white C WC W, 1:1
39. Baby 1 belongs to Mr. and Mrs. Jones. Baby 2 belongs
to the Guttierez family. Mr. and Mrs. Guttierez
cannot have a type O child as neither carries the
recessive allele.
40. If the father is not hemophiliac there is no chance of
a daughter having the disease. There is a 25 percent
chance that the son is a carrier since the mother is
a carrier.
41. Epigenetics studies the changes of inheritance of
certain traits based upon gene function, not DNA
sequence. Genetics focuses on DNA sequence. The
interaction between the environment and the genotype
can turn on or off certain phenotypes (fur colour in
Siamese cats, for example).
42. See Figures 4.7 (student textbook page 165) and
4.9 (student textbook page 167).
43. See Figure 4.13 on student textbook page 171.
44. Diagrams should show detail from the right side of
Figure 4.14 on student textbook page 172, with the
addition of details on anaphase I/telophase I,
and meiosis II. The first should indicate that
chromosomes look different as a result of crossing
over. For meiosis II, a note should indicate that
chromosomes go to daughter cells randomly as a result
of independent assortment.
45. Sample answer:
Cons—To increase fungal resistance, genes from other
plants or bacteria can be added to a crop to allow it to
encode enzymes which break down chitin or glucan,
respectively, which are essential components of fungal
cell walls.
Unless the crop is clearly labelled as being genetically
modified, there could be a danger of allergic reaction in
some people when they ingest food from the crop. (For
example, potatoes containing peanut genes may cause a
reaction in a person allergic to nuts.)
If the crop is near another crop, cross pollination of the
genes may also occur.
Pros—Fungi are responsible for a range of serious plant
diseases such as blight, grey mould, bunts, powdery
mildew, and downy mildew. Crops of all kinds often
suffer heavy losses. By genetically modifying these
crops, they will be more productive and reduce or
eliminate the need to use chemical fungicides or heavy
metals which are very damaging to the environment.
46. See Figure 4.29 on student textbook page 188.
47. Let T be tall, t be short. Each cross in the Punnett
square represents a 25 percent probability.
Parent genotypes: Tt
Punnett square: F1
T
t
T
TT
Tt
t
Tt
tt
Genotypic ratio of TT:Tt:tt is 1:2:1.
Phenotypic ratio of tall:short is 3:1.
Biology 11 Answer Key Unit 2 • MHR TR
25
48. Perform a test cross by crossing the unknown parent
with a homozygous recessive parent (tt). One short
offspring tells us that the parent is heterozygous:
T
?
t
TT
Tt
t
Tt
tt
No short offspring tells us the parent might not
be heterozygous:
T
T
t
TT
Tt
t
Tt
Tt
49. a. Answers should include: autosomal dominant;
brain deterioration over 15 years; decreased muscle
coordination; there is no cure; medications can
lessen symptoms. Diagnosis is done by looking for
CAG repeats in the gene called huntingtin.
b. Supporting answers might include:
Accuracy—35 or fewer repeats will not be affected
but 36–39 are at risk and over 39 will definitely get
the disease.
Uncertainty— the number of repeats may increase
or decrease between generations
Odds—due to the autosomal dominant nature, if
one parent has Huntington’s, there is at least a 50
percent chance that children will inherit the gene.
Knowing how severe the impact on the huntingtin
gene is will help plan the future.
50. See Figure 6.11 (student textbook page 252).
51. Dominance—Mendel’s peas exhibited dominance/
recessiveness. Tall is dominant over short, therefore
heterozygous plants were tall.
Incomplete dominance—Snapdragon flower colour
exhibits incomplete dominance: white and red are both
expressed and the heterozygote shows a blend of the
two alleles (it is pink). See Figure 6.1 (student textbook
page 242).
Codominance—both alleles are expressed. Roan cattle
show codominance in their hairs: some patches of fur
are white and other patches are red. The phenotype
is not a blend but is an intermediate. See Figure 6.2
(student textbook page 243).
Sex-linked inheritance—the allele is carried on the
portion of the X chromosome that does not have a
homolog on the Y chromosome, this results in more
males having characteristics such as colour deficient
vision. See Figure 6.13 (student textbook page 254).
26
MHR TR • Biology 11 Answer Key Unit 2
52. See Figure 6.14 (student textbook page 255) as a
reference. In this case, every letter B or b should be
an H or h. The key would be X HX H = normal female;
X HX h = carrier female; X hX h = hemophiliac female;
X HY = normal male; and X hY = hemophiliac male.
This pedigree has more affected males than females.
The affected characteristic can skip a generation (as
shown in generation II) but the next generation shows
only affected males.
53. Look for evidence that students have used knowledge
from this unit.
54. To try to make it so that the person can make their
own insulin. By gene therapy, it is hoped to correct the
defective gene.
55. Reduced genetic variability. Studies indicate that
animals born and raised in their native environment
do better than those that are imported.
56. a. Embryos used to be the only source of stem cells.
Now scientists can make specialized adult stem cells
into undifferentiated stem cells that can differentiate
into any of the three germ layers.
b. Sample answer: Dr. Lynn Megeney is focusing on
adult cardiac muscle stem cells. He is developing
small molecule compounds to stimulate the
regeneration of cardiac muscle tissue. This
will lead to being able to repair hearts disease
damaged hearts.
57. Sample answer: Ethically, we cannot force people
to breed just to study the genotypic ratios of
their offspring. Practically, we cannot control the
environmental factors that influence gene expression.
Scientifically, many genes are linked, some phenotypes
are polygenic and some phenotypes are not activated
unless certain environmental conditions are present.
58. a. As certain traits were bred into the breed, natural
variation was decreased. Perhaps by increasing the
jaw strength of a dog, you also decrease the hip
stability because the genes are linked. Some breeders
breed to close relatives, which increases the chance
for disorders to be exhibited.
b. Breed with dogs from different areas of the country
or world. Make sure that you are not breeding
close relatives.
59. a. CF is an autosomal recessive disorder. A defective
protein in the cell membrane leads to increased
mucus production.
b. By using a pedigree, a genetic counsellor can predict
whether either parent is a carrier and therefore
calculate the chance of having a child with CF.
c. Gene therapy could replace the gene (CFTR) on
chromosome 7. The challenge is that any one of 1600
mutations to the gene causes CF.
60. Answers should include use of a vector-like virus to
remove and replace the gene. It will be more accurate
if it is done in early development and only acts on a
simple mutation (unlike CF). A possible plot twists is
that the newly inserted gene could be from another
animal and carry with it other characteristics.
61. a. Sample answer: Thermal imaging or blood tests to
monitor protein production.
b. It reproduces asexually through budding and
therefore all offspring are identical (unless mutation
occurs). It can be transformed easily. It also
reproduces very quickly under optimum conditions
and is eukaryotic (therefore has large, non-coding
regions of DNA like human DNA does).
62. a. As data is collected from cancer patients, it can be
analyzed for similarities. DNA sequences can be
identified as markers to help predict cancer.
b. Students may name the Ontario Institute of
Cancer Research.
c. Refer students to BLM A-10 Presentation
Checklist or BLM A-32 Presentation Rubric (on
the accompanying CD) for assistance in preparing
presentations or assessing them.
63. Knowing your genetic profile may enable you to seek
treatment, change your lifestyle to reduce risk, or make
an informed decision on the risk of having children.
However, this knowledge could reduced your freedom,
create discrimination (in hiring, getting insurance,
or finding a life partner), and increase stress. These
implications show that society needs to consider how
the information can and should be used.
64. Sample answer: The ability to use adult stem cells to
create stem cells that can differentiate into any germ
layer has eliminated the need to use embryonic stem
cells (an ethical problem) for stem cell research that
could repair damaged heart and nerve tissues.
Unit 2 Self-Assessment Questions
(Student textbook pages 286–7)
1. d
2. d
3. e
4. d
5. b
6. c
7. b
8. c
9. b
10. c
11. Haploid cells that are genetically unique; reproductive
cells (sperm or egg)
12. The cells that result from meiosis merge during
fertilization to produce the first cell (zygote) of an
organism. This cell will carry all errors that occurred
during meiosis and is the foundation for the entire
organism. Mitosis begins and continues for the life of
the organism. Only the cells that divide from a cell that
has a mistake in mitosis will carry that error.
13. See the right-hand side of Figure 4.14 (student
textbook page 172).
14. See Table 4.2 (student textbook page 179).
15. Sample answer: Artificial insemination is used to
increase the availability of high-quality male semen to
breeders all over the world. This can be combined with
embryo transfer and genetic screening to ensure that
genetic disorders are not enhanced.
16. Rr and rr; R is round; The ratio of the offspring is close
to 1:1.
17. Ss and ss. The long-haired cat must be heterozygous to
have homozygous recessive offspring.
18. 25%
19. The pedigree indicates whether there is any chance of
a person or their child having a disorder. The genetic
testing can verify the findings of the counsellor. The
tests are expensive and should only be done to verify
the genotype of an individual.
Biology 11 Answer Key Unit 2 • MHR TR
27
20. Sample answer:
Pro
• improved quality of life
• lower stress on health–care
system because people are
healthier
• longer life
• cure for disease
• ability to repair tissues,
including missing limbs
• potential for designer
organisms
Con
• transgenics—should a
gene from one organism be
introduced into a another?
• increased stress on healthcare system because people
live longer
• virus may not find the correct
point in the DNA, leading to
more damage
• potential for designer babies
• “playing god”
• strong immune response to
the virus is possible
21. In the case of incomplete dominance, the offspring will
be grey (a blend of the two parental phenotypes). In
the case of codominance, the offspring will have black
and white patches.
22. The Rousseaus are parents of Baby 1; Baby 2 belongs to
the Sakic family.
23. a. Let X m = Duchenne muscular dystrophy allele
and let X M = normal allele; female carrier X M X m,
normal male X MY, affected male X mY
b. 25%
c. No. The father must have the recessive allele on
his X chromosome for him to father a daughter
with the disorder (she must be homozygous for the
allele). The father is normal and therefore has the
dominant allele.
24. Sample answer: By documenting the variations in
the human genome, scientists will have more data to
analyze and may find that some diseases are caused by
multiple genes or only certain environments.
25. The Human Genome Project determined the sequence
of the human genome (as well as other organisms).
This has lead to advancements in gene therapy (for
example, being able to identify a sequence that is faulty
and replace it), genetic testing (looking for faulty
sequences), and increased understanding of protein
action and treatments.
28
MHR TR • Biology 11 Answer Key Unit 2