Download SCI 30 UA Chapter Unit Review AnsKey

Unit A: Maintaining Health
Chapter and Unit Review Suggested Answers
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Science 30
Science 30 © 2007 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page.
Science 30
Unit A: Maintaining Health
Chapter and Unit Review Suggested Answers
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Knowledge
1. The four main jobs of the human circulatory system are to replace carbon dioxide and wastes within the
body’s cells with oxygen and nutrients to send chemical messengers throughout the body, to distribute body
heat, and to defend against disease.
2. This table compares the chambers of the mammalian heart.
Heart Chamber
Location
Type of Blood Found
in Chamber
right atrium
top right
deoxygenated
receives blood from
body from vena cava
right ventricle
bottom right
deoxygenated
receives blood from
right atrium and
delivers blood to lungs
left atrium
top left
oxygenated
receives blood from
lungs
left ventricle
bottom left
oxygenated
receives blood from
left atrium and delivers
blood to body via aorta
Function
3. Arteries and veins differ in the following ways:
• Arteries carry blood away from the heart, they usually carry oxygenated blood, and they have thicker
walls and are more elastic than veins.
• Veins carry blood toward the heart, they usually carry deoxygenated blood, they have thinner walls than
arteries, and they have one-way valves.
4. Plasma, red blood cells, white blood cells, and platelets is the order of blood components ranked by their
relative proportions in a sample.
5. The blood component responsible for initiating the clotting process is platelets.
6. Cardiovascular disease is one of many disorders that affect the heart and/or circulatory system. Examples
include coronary heart disease, strokes, and varicose veins.
7. Both a heart attack and a stroke are caused by thrombosis or clots. However, in a heart attack the coronary
arteries that supply the heart are blocked. This causes cell death in the heart. In the case of a stroke, a blockage
in the arteries that supply the brain causes the death of brain cells.
8. Disease-causing pathogens can enter the body through vectors, inhaled droplets from an infected person, cuts
or breaks in the skin barrier, and via food or water contaminated by pathogens.
Science 30: Unit A
Chapter and Unit Review Suggested Answers
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Chapter 1 Review Questions
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9. A vaccination involves the injection of dead or weakened disease-causing pathogens into body tissues by
using a hollow needle. With few adverse effects, the body produces an immune response to the weakened or
dead pathogens. This production of memory cells and antibodies helps the body respond quickly to future
exposure from live pathogens.
10. The answers to these questions can be checked by referring to “Overview of Immune Response” on page 65
of the textbook.
Applying Concepts
11. a. The greater the organism’s size, the lower the resting heart rate.
b. Since this is an estimate, answers may vary. A reasonable estimate for the resting heart rate of a 3-kg cat
should be about 175 beats per minute.
c. Since this is an estimate, answers may vary. A reasonable estimate for the resting heart rate of a
7000-kg T. rex should be about 28 beats per minute.
d. Since this is an estimate that is not bounded by two data points, answers may vary considerably to this
question. One possible answer is 20 beats per minute. This question demonstrates the limitations of using
data that goes beyond the end points of the data set.
e. The data provided is for mammals only. Extinct dinosaurs could have had different physiological
adaptations for their circulatory systems and, therefore, cannot be directly compared.
12. Evaluate your model according to the specific design criteria provided. All key structures on the heart
diagram are to be included in the model. Even though the model will not include labels, if someone points to
a heart structure on the model, you should be able to answer the following focusing questions:
• What is the name of this structure?
• Where would blood in this part of the heart go next?
• Does this part contain oxygen-rich blood or oxygen-poor blood?
13. Evaluate the models of other students according to the same design criteria used to evaluate your model.
14. a. Determine the strong points of your model by using the feedback from other students and teachers.
b. A great source of ideas for how to improve each model is evident in the work of the other students.
Science 30: Unit A
Chapter and Unit Review Suggested Answers
Hi, I’m Robbie the Red Blood Cell. My friend
Tina Toe Cell asked me to bring her some oxygen
and dispose of some waste. Here is my
ADVENTURE
The
blood re-enters the
heart from the lungs
through the pulmonary veins.
It then enters the left atrium.
The blood is then pumped into
the left ventricle. From the left
ventricle it goes into the aorta
where it starts travelling to the brain
and the entire body.
The blood first
enters into the heart
by the vena cavae and
enters the right atrium. The
blood is then pumped into
the right ventricle. From the
right ventricle the blood
enters the lungs through the
pulmonary artery.
The blood
then starts travelling throughout your
body down the main
artery. It branches off
into small capillaries
that carry oxygen-rich
blood to your entire
body while picking up
wastes and carbon
dioxide and bringing it
to disposable organs.
The blood
continues down the
leg, still with the
capillaries to feed the
muscles in your leg.
But it’s
not so
difficult
for your
body. It
is doing
it constantly.
Finally, the
blood reaches you. It
seems like a long process,
but it actually only takes the
body a matter of seconds for it
to be pumped so far away from
the heart. Without the heart, the
blood would not circulate the
body. This could result in limbs
that do not function properly.
Science 30: Unit A
So
that’s how
it all works.
Who could
know that it
could be such a
complicated
process?
Chapter and Unit Review Suggested Answers
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15. Although answers will vary, a sample comic strip is shown.
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16. a. Injecting more red blood cells creates an advantage for athletes because more red blood cells means they
have a greater oxygen carrying capacity. This would give the athletes a cardiovascular advantage.
b. It is more difficult to prove that an athlete has engaged in blood doping because drugs can be detected
in the blood, whereas blood itself is natural and it’s difficult to prove that extra blood cells are a result of
doping rather than high-altitude training.
c. Blood doping will most likely cause an increased density to the blood and, therefore, an increased
blood pressure.
d. Possible negative health effects of the practice of blood doping include an increased chance of infection
due to the injection and an increased blood pressure during exercise. Also, the removed blood may not be
stored properly in these illegal circumstances, resulting in dead or clotted blood being injected back into
the body.
e. Responses will vary for this question. You may argue that high-altitude training relies upon the efforts of
the athlete, while blood doping relies upon the expertise of technicians and medical staff. From this point
of view, blood doping may appear to be unethical.
17. Responses will vary somewhat. The activity can be scored using the scoring guide for skills shown in the
Science Data Booklet.
A sample experimental write-up follows. Your response should be similar.
Problem
Does the use of the manufacturer’s weight-loss drug cause an increased heart rate?
Hypothesis
If people take the drug, there is a greater incidence of increased heart rates.
Manipulated Variable
use of drug
Responding Variable
heart rate
Controlled Variables
Control the test groups for age, gender, amount of exercise, diet during the study, and heart rate prior to
the drug trial.
Science 30: Unit A
Chapter and Unit Review Suggested Answers
Measure and record the cardiac health of your test subjects by recording their heartbeats both at rest and
during exercise with a heart-rate monitor. Split your test subjects into two approximately equal groups.
Administer the drug to one group and give a placebo to the other group. The study should be done as a
double-blind test so that neither the researchers nor the subjects are aware of who is taking the placebo
and who is using the actual drug. At regular intervals over a period of six months, record the resting and
exercising heart rates of all participants. Participants should keep a log of any adverse effects. At the end
of the study, compare the reported and recorded effects from each of the two groups to determine if the
group administered the drug actually reports significantly more adverse effects than the placebo group.
18. You should be able to list some general adaptations to increase the ability of diving mammals to stay
underwater for prolonged periods. These include an increased lung capacity, a greater amount of blood or a
greater ability of blood to carry oxygen, more efficient lungs, and a slower use of oxygen while underwater.
The Weddell seal is able to remain underwater for more than an hour without surfacing to breathe because of
the following adaptations of the seal’s circulatory and respiratory systems:
• The seal has twice the volume of blood per kilogram as compared to a human, so it can store twice as
much oxygen.
• The seals have a diving reflex that, while underwater, slows the pulse and reduces oxygen consumption.
• Their muscles have a protein called myoglobin—similar to hemoglobin in blood—that stores oxygen in
muscles.
19. a. Patient 2 has high cholesterol, high blood pressure, and a high resting heart rate. These are all risk factors
for developing heart disease and its associated problems like a heart attack or a stroke. The doctor would
probably recommend that this patient stop smoking, exercise more, and follow a diet low in cholesterol
and saturated fats.
b. Patient 1 has a very high white blood cell count compared to the other two patients. This elevation can
indicate an infection.
c. Patient 3 does regular intensive exercise. This person’s heart is working very hard during this exercise,
which increases the heart’s capacity to stretch and expand. This means that to circulate the blood, the
more efficient heart of an athlete has to work much less when it’s at rest.
20. Students who have been actively focusing on questions 21 to 24 will have evidence of their active reading
strategies. This evidence can take the form of point-form notes or of highlighted passages on the print copy.
21. a. The first test done was simply the resting heart-rate value. The second test was the heart-rate increase,
which was measured by subtracting the resting heart rate from the peak exercise heart rate. The final test
was heart-rate recovery, which was the difference in heart rate between the peak value during exercise
and the value one minute after the exercise had stopped.
b. The test that appears to have the best ability to predict sudden death from a heart attack is heart-rate
increase. Participants with a heart-rate increase of less than 89 beats per minute had four times the risk of
sudden death due to heart attack compared to those with heart-rate increase values in excess of 113 beats
per minute.
Science 30: Unit A
Chapter and Unit Review Suggested Answers
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Experimental Design
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22. The subjects in the study had to meet the following criteria. They had to
• be men employed in the French civil service
• be between 42 and 53 years old
• be free of any known or suspected cardiovascular disease
• have a resting systolic blood pressure less than 180 mmHg
• complete the exercise testing
This screening procedure was used to determine if there was a test that could determine the likelihood
of sudden death from a heart attack in a population of men who seemed quite healthy and who had no
symptoms of cardiovascular disease.
These screening procedures were also important for the health of the participants—if they suffered from
cardiovascular illness or high blood pressure, the exercise test could have been dangerous for them.
23. a. The control group consisted of 5503 men. These subjects, during follow-up, were either alive or had died
from causes other than heart attacks.
b. There were 81 men from the original group who died suddenly due to heart attacks.
24. This research did not test a representative sample of the whole population. Only men who worked in the
French civil service were tested, so it would be difficult to extend the research results to the entire population
or to women since they were not part of this research.
Chapter 2 Review Questions
Knowledge
1. A chromosome is a coiled section of DNA visible using a microscope. DNA is the molecule that a
chromosome is made from. A gene is a region of DNA that provides instructions for making a protein.
Different genes are found in different locations on particular chromosomes. Note that chromosomes are only
visible with a light microscope during meiosis and mitosis.
2. In an autosomal cell, humans possess 23 pairs or a total of 46 chromosomes (22 pairs of autosomal
chromosomes and 1 pair of sex chromosomes).
3. Mitosis produces two genetically identical diploid daughter cells from one diploid body cell. Meiosis
produces four genetically different haploid cells from diploid cells located in the ovaries or the testes.
4. Acquired characteristics are traits developed or present as a result of environmental influences. An example
of an acquired characteristic is the ability to speak a particular language. Inherited characteristics include
traits present in an individual as a result of the genes (genotype) that the organism possesses. An example of
an inherited characteristic is a person’s eye colour.
5. a. Green pea pods are dominant since they appear to mask the yellow pea pod colour.
b. Letters used may vary. The parent pea plant will be GG for the green pod pea plant and gg for the yellow
pod pea plant. The offspring is heterozygous with a Gg genotype.
Science 30: Unit A
Chapter and Unit Review Suggested Answers
7. a. original strand:
complementary strand:
ATATACCAGCCGATA
TATATGGTCGGCTAT
ATA-TAC-CAG-CCG-ATA
Isoleucine-Tyrosine-Glutamine-Proline-Isoleucine
original strand:
amino acid sequence:
b. original strand:
complementary strand:
GCATGGTTCATAAGG
CGTACCAAGTATTCC
GCA-TGG-TTC-ATA-AGG
Alanine-Tryptophan-Phenylanine-Isoleucine-Arginine
original strand:
amino acid sequence:
c. original strand:
complementary strand:
CGTATGCCAGTTTAT
GCATACGGTCAAATA
CGT-ATG-CCA-GTT-TAT
Arginine-Methionine-Proline-Valine-Tyrosine
original strand:
amino acid sequence:
d. original strand:
complementary strand:
GGTTTATGCATTTCT
CCAAATACGTAAAGA
GGT-TTA-TGC-ATT-TCT
Glycine-Leucine-Cysteine-Isoleucine-Phenylanine
original strand:
amino acid sequence:
8. a. amino acid sequence:
Methionine-Threonine-Glutamine
possible DNA triplets:
ATG —
b. amino acid sequence:
possible DNA triplets:
Science 30: Unit A
ACT
ACC
ACA
ACG
CAA
—
CAG
Arginine-Lysine-Tryptophan
CGT
CGC
CGA
CGG
AGA
AGG
AAA
—
— TGG
AAG
Chapter and Unit Review Suggested Answers
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6. The term genotype refers to the alleles—the copies or forms of a particular gene—that an individual
possesses. Genotypes can be represented using letters for the different alleles. The term phenotype refers to
the physical characteristics or expression of the genotype. An example of a genotype is the listing of alleles
for the eye colour genotype or describing alleles as being homozygous or heterozygous for a trait. An example
of a phenotype is the eye colour of an individual.
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c. amino acid sequence:
possible DNA triplets:
d. amino acid sequence:
possible DNA triplets:
Serine-Proline-Aspartate
TCT
TCC
TCA
TCG
AGT
AGC
—
CCT
CCC
CCA
CCG
—
GAT
GAC
Leucine-Cysteine-Valine
TTA
TTG
CTT
CTC
CTA
CTG
—
TGT
TGC
—
GTT
GTC
GTA
GTG
9. Proteins are important and versatile molecules within a biological system. They form enzymes and
hormones, as well as structural or transport components of the cell, and they can be used as an energy source
for the body when carbohydrates and fats are unavailable.
10. Both types of mutations are changes in the DNA sequence, and they are often caused by the presence of
environmental mutagens. A point mutation is the substitution of a nucleotide in the DNA sequence, while a
frameshift mutation is an addition or deletion of a nucleotide. Both types of mutations change the sequence
of amino acids in a gene.
A significant difference in these types of mutations is that a point mutation alters one nucleotide base, which
may cause a change in one amino acid in a sequence. Since a frameshift mutation literally shifts the frame
or the grouping of all the nucleotides “downstream” from the substitution or deletion, this type tends to be a
more damaging form of mutation.
11. The steps involved in creating recombinant DNA are as follows:
(a) A gene is located on a chromosome.
(b) Enzymes are used to cut the chromosome’s DNA into pieces, and the desired gene is isolated.
(c) The segment of DNA containing the desired gene is inserted into a bacterial plasmid. The plasmid
containing the genes from two or more organisms is called recombinant DNA.
(d) A recombinant DNA plasmid can be inserted into a bacterium where it will be replicated during cell
division. Copies will appear in subsequent daughter cells.
Science 30: Unit A
10
Chapter and Unit Review Suggested Answers
Comparing Mitosis and Meiosis—Labelled
12.
Mitosis
Meiosis
DNA replicates
DNA replicates
pairing of
homologous
chromosomes
alignment along centre
separation of chromosomes
chromosome
exchange of
genetic segments
between homologous
chromosomes
(crossing over)
separation of chromosomes
cell division
cell division
daughter cells: 2n
cell division
to form gametes
gametes: 1n
Science 30: Unit A
11
Chapter and Unit Review Suggested Answers
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Applying Concepts
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13. a. Since the gene for white colour is recessive, breeding a white tiger with a standard-coloured tiger would
most likely produce offspring without the white phenotype. The standard-coloured tiger’s dominant allele
masks the recessive white gene in the offspring.
b. The inbreeding of any population often results in an increase in the frequency of certain traits. In white
tigers, inbreeding could reduce the variety of alleles that could be inherited. Inbreeding could result in
more offspring developing autosomal recessive genetic diseases if one of the parents possessed such an
allele.
14. a. Since there was no family history of this trait, the red fur is most likely the result of a spontaneous mutation.
b. It can be determined if the new colour trait is dominant or recessive by breeding the puppy with the red
fur trait with another dog and observing the phenotypes of the offspring. If some or all of the offspring
have red fur, then the trait is most likely dominant. If no offspring has the trait, then red fur is most likely
recessive.
A dog with red fur—the dominant trait of RR or Rr—is crossed with a dog with other-coloured fur (rr).
R
R
r
Rr
Rr
r
Rr
Rr
R
r
r
Rr
r r
r
Rr
r r
Either all offspring would have the red-fur trait or half the offspring would have the trait.
A dog with red fur—the recessive trait of rr—is crossed with a dog with other-coloured fur (RR).
r
r
R
Rr
Rr
R
Rr
Rr
All offspring have the other-coloured fur phenotype.
c. Ideally, the breeder should find another dog with a similar fur colour phenotype and breed them together.
If offspring are only bred that have red colour fur (since they have at least one red fur-colour allele), there
is a better chance of developing offspring that are homozygous for that red fur-colour trait.
15. Antibacterial soaps may kill most of the bacteria. The few bacteria with some natural resistance to these
soaps are the ones that survive to reproduce. These somewhat resistant bacteria can transfer plasmids
between other bacteria sharing the genes that cause antibacterial soap resistance. Eventually, a strain of
antibacterial soap-resistant bacteria may develop.
Science 30: Unit A
12
Chapter and Unit Review Suggested Answers
Gene Therapy for Cystic Fibrosis
tissue cell
nucleus
healthy person
genetic
material
modification to viral DNA
so it cannot cause disease
modified adenovirus
(cold virus)
chromosome 7
“healthy” gene:
codes for proper amino
acid sequence—no
cystic fibrosis symptoms
healthy gene
is spliced into
adenovirus
cells lining lung
tissue of person
with cystic fibrosis
adenovirus with
healthy gene goes
to lung tissue
adenovirus delivers
healthy gene:
now cell has code for proper
amino acid sequence—if
successful, symptoms of
cystic fibrosis relieved
person with
cystic fibrosis
Science 30: Unit A
13
Chapter and Unit Review Suggested Answers
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16.
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17. a. Agree. Since there are three phenotypes, this trait cannot be controlled by one gene with two alleles—
this could indicate co-dominance.
b. Disagree. If they all had homozygous genotypes, the greatest number of phenotypes of the offspring
could be two—the homozygous recessive colour and the homozygous dominant colour.
c. Disagree. One or both of the parent dogs are carriers for a different allele or possess one of each allele
type to create a different phenotype.
d. Agree. The phenotype provides some information about an organism’s genotype.
18. a. Only males are affected by a gene present on the Y chromosome.
b. Since males get only one Y chromosome, they express all the alleles on that chromosome. There is no
possibility of dominance masking another allele because there is no other allele to mask.
c. The percentage probability of a hairy-eared man passing this trait onto his son is 100%. Sons always
inherit the Y chromosome from their fathers, since the inheritance of the Y chromosome is what makes a
child male.
d. The percentage probability of a hairy-eared man passing this trait onto his daughter is 0%. Daughters
never inherit the Y chromosome. The inheritance of the Y chromosome is what makes a child male.
e. No. Carriers have two copies of a gene, and the dominant allele masks the expression of another
recessive allele. If an individual inherits the Y chromosome, he will be male and express all the traits of
that chromosome.
19. a. The pedigree indicates that the inheritance of the disease follows an autosomal-dominant inheritance
pattern. Since no one gender is more affected than the other, it does not appear to follow a sex-linked
pattern. The occurrence of the genotype does not skip generations, so it is most likely not recessive.
b. The cross of the two parents in generation IV can result in two genotypes and phenotypes. It is important
to remember that since this allele appears to be dominant, the affected male can have either of the
genotypes EE (homozygous dominant) or Ee (heterozygous dominant). The female parent is most likely
heterozygous for the trait.
If the male is homozygous, then all offspring will be EE or Ee and all will show symptoms of the
disorder.
If the male is heterozygous, then all offspring will be EE, Ee, or ee. The percentage probability of the
offspring having the disorder is 75%, and the percentage probability of the offspring either not showing
the symptoms or not being carriers of the affected allele is 25%.
Science 30: Unit A
14
Chapter and Unit Review Suggested Answers
1.
a.
c.
e.
g.
2.
3.
b.
d.
f.
h.
atrium
septum
vena cava
pulmonary vein
ventricle
heart valves
pulmonary artery
aorta
Artery
Vein
Capillary
Sketch: should show a
thick-walled vessel
A red/pink colour can be used
to illustrate the oxygenated
blood flowing through the
vessel.
Sketch: should show a
thin-walled vessel that should
include valves
A blue/purple colour can be
used to illustrate deoxygenated
blood flowing through the
vessel.
Sketch: should show a tiny
thin-walled vessel only one cell
thick
Role: to carry oxygen-rich
blood away from the heart;
(exception: pulmonary artery)
Role: to carry oxygen-poor
blood toward the heart;
(exception: pulmonary vein)
Role: thin enough to allow gas
exchange between cells and
bloodstream
a. b. c. d. white blood cell
sample of plasma
red blood cell
platelet
4. Similarities and differences between anginas, heart attacks, and strokes are shown in this table.
Similarities
Differences
• The chances of developing all three disorders
can be increased by stress, smoking, a lack
of exercise, and a diet high in cholesterol and
saturated fats.
• A heart attack results in damage to the muscle
cells of the heart. It is caused by a blockage in
an artery that delivers oxygenated blood to the
heart.
• Heredity can play a role in developing any one of
these conditions.
•A stroke is caused by a blockage in the arteries
that supply blood to the brain.
• A buildup of plaque in arteries can play a factor
in developing these conditions.
•A constriction of the coronary vessels of the
heart causes angina, while a complete blockage
of these vessels causes a heart attack.
• Anginas and heart attacks affect the coronary
vessels of the heart.
• Both a heart attack and a stroke can be caused
by a blood clot (thrombosis).
Science 30: Unit A
15
• Angina is a pain caused by oxygen-starved
heart muscles but both strokes and heart
attacks cause cells to die.
Chapter and Unit Review Suggested Answers
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Unit A Review Questions
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5. The following table matches each type of white blood cell to a person or an object from a hockey game.
Person or Object and Role
6.
Type of White Blood Cell
Coach—co-ordinates and directs team’s moves
and analyzes other team’s plays
d. helper T-cells
Offensive players—try to keep opposing team’s
players and puck on opposition side
a. macrophage
Coach’s playbook—keeps records of strategies
used to win games
b. memory B- and T-cells
Referee—calls an end to game
e. suppressor T-cells
Jersey—makes a distinction between players
f. antigen
Defensive players—respond to opposing players
who have broken through their defensive line
c. B-cells
a.
b.
c.
d.
e.
f.
g.
h.
karyotype
gamete
DNA
gene
amino acid
chromosome
protein
DNA base triplet
7. a. WW or Ww
c. W or w b. ww
d. vestigial wings
8. Alternating deoxyribose sugars and phosphate groups form the spine or each side of the DNA ladder.
Attached to each of the sugar groups is one of four nitrogen bases—adenine, cytosine, guanine, or thymine.
The adenine bases pair with the thymine bases, and the cytosine bases pair with the guanine bases. These
pairings join together the two complementary halves or strands of the DNA molecule. Students may use a
diagram to explain the structure of DNA.
9. a. Autosomal genetic diseases are caused by non-functioning genes found on one of the 22 pairs of
autosomal chromosomes. Both males and females have two copies of the gene that causes the genetic
disease. If the disease is autosomal dominant, offspring need only one copy of the improperly functioning
gene to develop the genetic disease.
b. If the disease is autosomal recessive, then offspring lacking a dominant functioning gene will develop
the disease.
c. Sex-linked genetic diseases are caused by non-functioning genes found on X or Y chromosomes.
Sex-linked diseases affect the two genders in different proportions. For example, genetic diseases caused
by genes on the X chromosome affect males more frequently than females because males have only one
copy of this gene.
Science 30: Unit A
16
Chapter and Unit Review Suggested Answers
step 1: Extract strands of DNA from cells in biological evidence.
step 2: Use enzymes to cut the DNA into smaller fragments.
step 3: Put the DNA fragments in a special gel placed within an electric field.
step 4: DNA fragments become separated as they are pulled through the gel by the electric field.
step 5: Observe and photograph the pattern of bands that form.
step 6: Perform the same procedure on DNA from the cells of suspects.
step 7: Compare the DNA fingerprint from the crime scene evidence to fingerprints from suspects.
step 8: A matching fingerprint will link the suspect to the biological evidence.
11. Oxygenated blood appears bright red while deoxygenated blood appears more bluish. Since less of the baby’s
blood is getting oxygenated, the blood in its vessels—including those capillaries near the skin’s surface—
would be bluer. This lack of oxygenated blood gives the baby’s skin a bluer appearance than usual.
12. a. The following Punnett square illustrates a cross between a male with the genotype Rh+/Rh- and a female
with the genotype Rh-/Rh-.
Rh–
Rh–
Rh+ Rh+/Rh– Rh+/Rh–
Rh– Rh–/Rh– Rh–/Rh–
b. The percentage probability that the offspring will have the Rh positive blood type is 50%.
c. The percentage probability that the offspring will have the Rh negative blood type is 50%.
d. The percentage probability that the offspring will have the Rh positive blood type but carry the
Rh negative blood type allele is 50%.
e. The Rh positive antigens on the surface of the red blood cells would not be recognized by the transfusion
recipient’s immune system. These donated blood cells would be treated as an invading pathogen. The
body would produce antibodies against these antigens to cause clumping of the blood. This response
could result in the death of the recipient who has received this non-compatible blood.
Science 30: Unit A
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Chapter and Unit Review Suggested Answers
Science 30 © 2007 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page.
10. The basic steps involved in producing and using a DNA fingerprint to positively identify biological evidence
left at a crime scene are as follows.
Science 30 © 2007 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page.
13. A patient will likely experience the symptoms noted given each of the following blood results.
Blood Test Results
Possible Symptoms
a low red blood cell count
The patient will most likely experience difficulty getting
sufficient oxygen and may feel short of breath.
a high white blood cell count
The patient is most likely battling an infection against
a disease-causing pathogen. The patient may feel the
symptoms of a cold or flu such as tiredness, fever, and
swollen lymph glands.
a very low T-cell count
The patient will have trouble fighting off infections and may
be sick with a rare disease or illness. The patient may have
contracted HIV, which destroys T-cells.
a low platelet count
The patient will have difficulty clotting blood when blood
vessels are ruptured, which results in excessive bleeding.
This may be a result of hemophilia.
14. a. The gene that produces these antigens in humans will be isolated and inserted into the DNA of a pig
embryo by the use of a modified virus or a bacterial plasmid.
b.
Risk
Benefit
Diseases that previously couldn’t be passed
are now passed from pigs to humans.
People would have a much shorter wait for
organs.
15. a. A negative effect from taking an immune-suppressant drug—such as cyclosporin—is the suppression of
the patient’s immune system, which makes the patient more susceptible to infectious diseases.
b. Identical twins have identical genetics so they have the same antigens on their organs. The organ
recipient’s immune system will not treat the new organ as an invading organism.
16. A karyotype can be used to determine what species the genetic sample was taken from (different species
have different numbers of chromosomes). This aspect of karyotyping can be used in wildlife forensics.
A karyotype can also determine the gender of an individual. This data can be used in forensic investigations.
And a karyotype can determine if an individual has missing chromosomes or extra chromosomes. This
information can be used in prenatal tests to determine if a baby has a genetic disorder such as Down syndrome.
Science 30: Unit A
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Chapter and Unit Review Suggested Answers
b. The offspring will be heterozygotes and possess both a dominant allele and a recessive allele.
c. His results are producing the same results as Mendel’s investigation of pea plants. Both of the parent
plants must be heterozygotic. If a Punnett square cross between two heterozygotic parents (Rr ¥ Rr) is
performed, the results illustrate a distinctive 3:1 ratio of the dominant trait over the recessive trait.
R
r
R
RR
Rr
r
Rr
r r
18. The point mutation in mutated sequence #2 changed a DNA triplet code that produced leucine to a “stop”
or “terminate” instruction. The amino acid chain would be incomplete and the protein would most likely not
function properly.
19. a. The following Punnett square illustrates the cross of alleles of Tsar Nicholas II and Tsarina Alexandra.
XH
Xh
XH
XH XH
XH Xh
Y
XH Y
Xh Y
b. The percentage probability of a male offspring from this cross developing hemophilia is 50%.
c. The percentage probability that a female will develop hemophilia from a cross between parents with
those genotypes is 0%. Alexei’s sisters could only have been carriers.
d. The non-functioning gene that caused Alexei’s hemophilia was present in every one of his cells.
This genetic defect made him unable to efficiently clot his blood, and there is no cure for this condition.
Alexei could have been treated only by repeated transfusions of blood or with a blood-clotting factor
from a suitable donor.
e. Each person has a unique sequence of bases in his or her DNA. A DNA fingerprint can be done to create
a pattern of bands unique to each individual. These DNA fingerprints can be used to positively identify
someone. More closely related individuals have more similar DNA.
20. a. The egg cell is haploid (1n) with one-half of the number of chromosomes as an autosomal body cell.
b. The autosomal cell would be diploid (2n) with two copies of each chromosome.
c. The clone zygote would be diploid (2n) like the body cell.
d. If the chromosomes weren’t removed, the fused zygote would have more chromosomes than needed.
The zygote would not develop properly.
Science 30: Unit A
19
Chapter and Unit Review Suggested Answers
Science 30 © 2007 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page.
17. a. Red is the dominant colour phenotype.
Science 30 © 2007 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page.
e. The processes of cloning and fertilization are compared and contrasted in this table.
Similarities
Differences
• With a clone, all the genetic information
comes from one individual.
• Both result in a diploid zygote.
• Both require implantation in a uterus to
develop completely. Development is the
same for both processes once the zygote
is implanted.
• The offspring produced by fertilization are
genetically unique. Clones are genetically
identical to their donors.
• Both result in a baby being produced—
cloning does not produce adults or
organisms with the memories of their
genetic donors.
• Two gametes are required for fertilization,
but only one egg cell and one body cell are
needed for cloning.
• Developing a clone zygote requires a process
carried out in artificial lab conditions.
f. Acquired characteristics shape an individual. Differences in experiences could create a different
personality or mannerism in the sheep. Differences in physical environmental factors, such as diet or
accidents, could cause physical differences between the clone and the original sheep.
Science 30: Unit A
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Chapter and Unit Review Suggested Answers