ANSWER KEY
Chapter 1
Chapter 1 Project Worksheet 1
(p. 6)
Students’ data tables and graphs will vary somewhat. In general, eggs will increase in size when
they soak in vinegar and plain water, and they will
decrease in size when they soak in liquids that have
a lower concentration of water, such as salt water.
Chapter 1 Project Worksheet 2
(p. 7)
Column 4 in the table should read: 70 percent (liquid A); 92 percent (liquid B); 87 percent (liquid
C); 79 percent (liquid D)
1. Liquids A and D would cause the cell to lose
water; liquids B and C would cause the cell to
take in water.
2. Possible answers might include shampoo,
syrup, ketchup, or honey.
110
Unit 1 Resources
Section 1-2 Review and Reinforce
(p. 15)
1. cytoplasm
2. endoplasmic reticulum
3. nucleus
4. mitochondrion
5. cell membrane
6. Organelles
7. cell wall
8. cell membrane
9. nucleus
10. chromatin
11. cytoplasm
12. Mitochondria
13. endoplasmic reticulum
14. Ribosomes
15. Golgi bodies
16. chloroplasts
17. vacuole
18. Lysosomes
Section 1-2 Enrich
(p. 16)
1. (1) builds new structures; (2) carries materials
from place to place; (3) produces power; (4) produces food; (5) disposes of waste; (6) controls
the rest of Cell City; (7) stores foods and water;
(8) controls what enters and leaves Cell City
Science Explorer Focus on Life Science
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Section 1-1 Review and Reinforce
(p. 11)
1. Hooke
2. First person to observe bacteria
3. Schleiden
4. Concluded that all animals are made up of cells
5. Virchow
6. Magnification and resolution are both important properties of a microscope. Magnification
is the ability to increase how large an object
appears, whereas resolution is the ability to
bring the details of the object into focus.
7. An electron microscope uses electrons instead
of light to magnify an object, and it can view
an object with much greater magnification
and resolution than a light microscope.
8. Cells are the basic unit of structure and function in living things, and all living things are
composed of cells.
9. b
10. e
11. d
12. c
13. a
14. g
15. f
Section 1-1 Enrich
(p. 12)
1. Transmission electron microscopes (TEM)
and transmission positron microscopes (TPM)
both use beams of atomic particles to view
specimens. However, the TPM uses beams of
positrons, which do not harm living things,
whereas the TEM uses beams of electrons,
which living things cannot withstand.
Therefore, unlike the TEM, the TPM can be
used to view living specimens.
2. Acoustic microscopes bounce high-frequency
sound waves off an object. The echoes of the
sound waves are then translated onto a screen
as a microscopic image.
3. Transmission positron microscopes and
acoustic microscopes can be used to view cells
that are still alive and functioning. Electron
microscopes, on the other hand, can be used
to view only cells that are no longer alive.
ANSWER KEY
2. (1) ribosome; (2) endoplasmic reticulum;
(3) mitochondrion; (4) chloroplast; (5) lysosome; (6) nucleus; (7) vacuole; (8) cell wall or
cell membrane
3. Cell City represents a plant cell because it contains a chloroplast.
Section 1-3 Review and Reinforce
(p. 19)
1. Sugars (or Starches)
2. Lipids
3. provide energy
4. Proteins
5. Nucleic acids
6. help produce proteins (or pass genetic material from parent to offspring)
7. l
8. k
9. i
10. g
11. a
12. c
13. f
14. h
15. j
16. e
17. d
18. b
© Prentice-Hall, Inc.
Section 1-3 Enrich
(p. 20)
Amino Acids
A
B
C
D
E
A
AA
AB
AC
AD
AE
B
BA
BB
BC
BD
BE
C
CA
CB
CC
CD
CE
D
DA
DB
DC
DD
DE
E
EA
EB
EC
ED
EE
1. Each letter pair represents a unique twoamino acid protein.
2. 25 (5 × 5)
3. 6 × 6, or 36; 20 × 20, or 400
4. Increasing the number of amino acids each
protein contains greatly increases the number
of unique proteins that could be formed from
just a few amino acids.
Science Explorer Focus on Life Science
Section 1-4 Review and Reinforce
(p. 23)
1. osmosis
2. diffusion
3. active transport
4. Diffusion is the passive transport of any molecules across a selectively permeable membrane, whereas osmosis is the passive transport
of water molecules across a selectively permeable membrane.
5. Both active and passive transport refer to the
movement of substances across a selectively
permeable membrane. Active transport
requires energy, whereas passive transport
does not.
6. Two methods of active transport are the use of
transport proteins and transport by engulfing.
7. Small size results in substances that enter and
leave the cell having just a short distance to
travel through the cytoplasm from the cell
membrane to the places where they are needed
within the cell.
8. true
9. Diffusion
10. osmosis
11. passive transport
12. Active transport
Section 1-4 Enrich
(p. 24)
1. Energy is not required because the passenger
molecule is moving from an area of higher concentration to an area of lower concentration.
2. Passenger molecules need to be helped because
they are unable to pass through the cell membrane on their own.
3. Active transport would be required because
energy would be needed to move the substance from an area of lower to an area of
higher concentration.
4. Facilitated diffusion with the help of a carrier
molecule does not require energy, whereas
active transport with the help of a transport
protein does require energy.
5. The person’s cells would be unable to take in
or release the substance because the substance
would be not be able to pass through the cell
membrane.
Unit 1 Resources
111
ANSWER KEY
Chapter 1 Skills Lab
(pp. 25–26)
For answers, see Teacher’s Edition, p. 22.
Chapter 1 Real-World Lab
(pp. 27–29)
For answers, see Teacher’s Edition, pp. 28–29.
Chapter 2
Chapter 2 Project Worksheet 1
(p. 34)
Students’ data tables should be similar to the
sample data table below.
Measures of Plant Growth
Date
and Time
Plant ID
Number/ Lighting
Letter Conditions
Plant
Height
(mm)
Plant
Diameter
(mm)
Number
of
Leaves
High
130
100
15
Looks healthy
Comments
October 1, 2:00 PM
1
October 1, 2:00 PM
2
Low
134
124
18
Looks healthy
October 3, 2:00 PM
1
High
132
120
17
Very green
October 3, 2:00 PM
2
Low
134
124
18
A little pale
Section 2-1 Review and Reinforce
(p. 39)
1–2. Carbon dioxide, Water
3–4. Glucose, Oxygen
5. Carbon dioxide and water
6. Oxygen and sugars, including glucose
7. Because sunlight provides the energy for the
reaction but is neither a raw material nor a
product of the reaction
8. In the chloroplasts of plants and some other
organisms, such as algae and some bacteria
9. photosynthesis
112
Unit 1 Resources
Pigments
chlorophyll
Stomata
autotroph
heterotroph
Section 2-1 Enrich
(p. 40)
1. Green; about 95 percent
2. Blue; about 90 percent
3. You see green, yellow, and orange.
4. You do not see violet, blue, or red.
5. Chlorophyll makes a leaf look green because it
reflects the highest percentage of green light.
Section 2-2 Review and Reinforce
(p. 43)
1–2. glucose, oxygen
3–5. carbon dioxide, water, energy
6. In the cytoplasm
7. In the mitochondria
8. Fermentation does not require oxygen, while
respiration does. It produces less energy than
respiration.
9. Alcoholic fermentation
10. Lactic-acid fermentation
11. respiration
12. Fermentation
Section 2-2 Enrich
(p. 44)
1. Students’ time lines should include the following events: People use fermentation to make
bread rise and produce alcoholic beverages
(8000 B.C.); Chinese use fermented soybeans to
treat skin infections (2000 B.C.); Chinese use
fermented tea to treat several illnesses
(220 B.C.); Schwann concludes fermentation is
due to living things (1840); Pasteur finds
fermentation is caused by yeast (1854); Buchner
receives Nobel Prize for showing yeast enzymes
cause fermentation (1907); Harden and EulerChelpin receive Nobel Prize for determining
how enzymes cause fermentation (1929);
fermentation produces antibiotics (1940s).
2. Louis Pasteur determined that the process of
fermentation is caused by yeast.
3. Two of the oldest uses of fermentation are to
make alcoholic beverages and to make bread rise.
Science Explorer Focus on Life Science
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Chapter 2 Project Worksheet 2
(p. 35)
1. Sunlight, water, and carbon dioxide
2. The production of sugars (food) by the plant
3. It makes the plant get bigger, or grow.
4. The plant would remain the same size, because
it would not have the food it needed for
growth.
5. The plant in the bright light will undergo
many hours of photosynthesis and grow
noticeably. The plant in the low light will
undergo very little photosynthesis and grow
little if at all.
10
11.
12.
13.
14.
ANSWER KEY
4. Fermentation is used in medicine today to
produce antibiotics.
© Prentice-Hall, Inc.
Section 2-3 Review and Reinforce
(p. 47)
1. Chromatin condenses
2. Metaphase
3. Chromatids separate
4. Telophase
5. interphase
6. interphase
7. cytokinesis
8. c
9. a
10 f
11. b
12. g
13. d
14. e
Section 2-3 Enrich
(p. 48)
1. The furrow forms when the cell membrane
pinches in around the middle of the cell. It
eventually divides the cell in two.
2. The stembody forms when the spindle fibers
are pressed together by the furrow. It is cut in
two when the cell divides.
3. Vesicles are pockets of cell-wall material. They
develop into new cell walls and cell membranes.
4. An animal cell would appear pinched in
around the middle by the cell membrane. A
plant cell would not be pinched in but would
have small structures (vesicles) lined up across
the middle of the cell.
Section 2-4 Review and Reinforce
(p. 51)
1. Mutations
2. Tumor
3–5. Surgery, Radiation, Chemotherapy
6. They disrupt the normal cell cycle and cause
cells to divide in an uncontrolled way.
7. Some of the cancerous cells may break off the
tumor, enter the bloodstream, and travel to
other parts of the body.
8. The risk of some types of cancer can be
reduced by not smoking, by eating a low-fat
Science Explorer Focus on Life Science
9.
10.
11.
12.
diet, and by eating a lot of fruits, vegetables,
and grain products.
Cancer
mutation
tumor
chemotherapy
Section 2-4 Enrich
(p. 52)
1. For males there were declines in cancer death
rates for age groups 35–44 and 45–54.
2. For females there were declines in cancer death
rates for age groups 25–34, 35–44, and 45–54.
3. males
4. Males aged 85 years and older
5. For 1980 the percent was 2,370/100,000 =
0.02, or 2%. For 1990 the percent was
2,740/100,000 = 0.03, or 3%.
Chapter 2 Real-World Lab
(pp. 53–54)
For answers, see Teacher’s Edition, p. 50.
Chapter 2 Skills Lab
(pp. 55–57)
For answers, see Teacher’s Edition, p. 59.
Chapter 3
Section 3–1 Review and Reinforce
(p. 67)
1. stem height or stem length
2. The two alleles are tall stems and short stems.
3. The dominant allele is tall stems because the
trait always shows up when the allele is present.
4. The recessive allele is short stems because it is
masked, or covered up, by the allele for tall
stems.
5. The F1 offspring have one allele for tall stems
and one allele for short stems. They received
the allele for tall stems from the tall parent and
the allele for short stems from the short parent.
6. g
7. f
8. d
9. h
10. c
11. b
Unit 1 Resources
113
ANSWER KEY
12. a
13. e
Section 3–1 Enrich
(p. 68)
1. The long-haired cat is a purebred because it has
two recessive alleles for long hair. If it were a
hybrid, it would have both the dominant and
recessive allele and have short hair because the
dominant allele always masks the recessive allele.
2. The short-haired cat is a hybrid because it had
offspring with long hair. The offspring receive
one allele from each parent, and in order to
have long hair, the offspring must receive a
recessive allele from each parent.
3. All offspring would have short hair, none
would have long hair.
4. The black horse is a hybrid because it produced an offspring with the trait controlled by
a recessive allele. In order to have the trait controlled by a recessive allele, the offspring must
receive a recessive allele from each parent.
5. Cross the smooth-coated guinea pig with a
rough-coated guinea pig. If any of the offspring have rough coats, then the smoothcoated parent is a hybrid. If all the offspring
have smooth coats, then the smooth-coated
parent is a purebred.
Section 3–2 Review and Reinforce
(p. 71)
1.
B
b
BB
Bb
b
Bb
bb
B
b
b
Bb
bb
b
Bb
bb
2.
3. The probability of an offspring being black is 3
in 4, 3⁄4, or 75 percent. The probability of an offspring being white is 1 in 4, 1⁄4, or 25 percent.
114
Unit 1 Resources
Section 3–2 Enrich
(p. 72)
1. All possible allele combinations that each parent
can pass on to the offspring are TG, Tg, tG, tg.
2. The possible genotypes are TTGG, TTGg,
TtGG, TtGg, TTgg, Ttgg, ttGG, ttGg, ttgg. The
possible phenotypes are tall with green pods,
tall with yellow pods, short with green pods,
short with yellow pods.
3. The probability of tall plants with green pods
is 9 out of 16, or 56 percent. The probability of
short plants with yellow pods is 1 out of 16, or
6 percent.
Section 3–3 Review and Reinforce
(p. 75)
1. Meiosis II
2. Meiosis I
3. End
4. Beginning
5. Meiosis II
6. Genes are carried from parents to their offspring on chromosomes.
7. When the sex cells combine, the offspring will
have exactly the same number of chromosomes in its body cells as did each of its parents. If sex cells had the same number of chro-
Science Explorer Focus on Life Science
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B
4. Yes, the principles of probability predict what
will likely occur, not what will occur.
5. One guinea pig parent is white, and the other
is black.
6. The white guinea pig parent is homozygous
(bb), and the black guinea pig parent is heterozygous (Bb). If the black guinea pig parent
were homozygous (BB), then there would be
no chance for the offspring to be white; they
would all be heterozygous black.
7. The probability of an offspring being black is 2
in 4, which is 1⁄2 or 50 percent. The probability
of an offspring being white is also 2 in 4, 1⁄2, or
50 percent.
8. f
9. a
10. e
11. g
12. b
13. c
14. d
ANSWER KEY
mosomes as body cells, then the offspring
would have two times the number of chromosomes in its cells as did its parents.
8. meiosis
© Prentice-Hall, Inc.
Section 3–3 Enrich
(p. 76)
Students’ models should be similar to the steps of
meiosis shown in Exploring Meiosis on page 99. Be
sure that each student ends with four sex cells with
half the number of chromosomes as the parent cell.
1. Meiosis is the process by which the number of
chromosomes is reduced by half to form sex
cells—sperm and eggs.
2. Before meiosis can begin, every chromosome
in the cell must be copied.
3. The chromosome pairs line up next to each
other, and then they separate and move to
opposite ends of the cell. The cell divides, and
each new cell contains one double-stranded
chromosome from each pair.
4. The double-stranded chromosomes line up in
the center of the cell, split apart, and move to
opposite ends of the cell.
5. Each sex cell has only half the number of
chromosomes that the parent cell had. This is
important because when sex cells combine to
produce offspring, each sex cell contributes
half the normal number of chromosomes.
Each offspring ends up with the normal number of chromosomes, not double the number.
6. Genes, the factors that control traits, are carried from parents to their offspring on chromosomes.
Section 3–4 Review and Reinforce
(p. 79)
1. neutral
2. harmful
3. helpful
4. harmful
5. Messenger RNA copies the coded message
from the DNA in the nucleus and carries the
message into the cytoplasm. In the cytoplasm,
ribosomes attach to the messenger RNA and
begin to “read” the three-letter code of bases.
Transfer RNA carries specific amino acids to
the ribosome, where they are added to the
growing protein chain. The protein is complete
Science Explorer Focus on Life Science
when the ribosome comes to a code that acts as
a stop sign and releases the messenger RNA.
6. The order of DNA bases along a gene forms a
code that specifies the order in which amino
acids will be put together to produce a protein.
7. transfer RNA
8. Messenger RNA
Section 3–4 Enrich
(p. 80)
1. The messenger RNA sequence is
AUGAAUGGCUCGAUCUGA.
2. The sequence of amino acids is methionine,
asparagine, glycine, serine, isoleucine, and “stop.”
3. The anticodon sequences for each of the transfer
RNA molecules for the amino acids in the
sequence are UAC, UUA, CCG, AGC, UAG, ACU.
4. The mutated sequence of amino acids is
methionine, glutamine, tryptophan, leucine,
aspartic acid, and leucine. This sequence of
amino acids is completely different from the
normal sequence, so the protein would not
function normally at all.
5. The mutated sequence of amino acids would be
the same as the normal sequence of amino acids
because the changed codon in the DNA still
specifies the amino acid glycine. The protein
would still have its normal function, and the
mutation would have no affect on the organism.
Chapter 3 Skills Lab
(pp. 81–82)
For answers, see Teacher’s Edition, pp. 76–77.
Chapter 3 Skills Lab
(pp. 83–85)
For answers, see Teacher’s Edition, pp. 84–85.
Chapter 4
Chapter 4 Project Worksheet 1
(p. 90)
1. Irene’s and Leo’s son is Gordon; their sons-inlaw are Richard and Tim.
2. Irene and Leo have 6 grandchildren; 3 of them
are girls.
3. Ralph’s father is Gordon; Ashley’s mother is
Cheryl.
4. Emily’s son is Richard, Jr.; Tim’s son is Zack.
Unit 1 Resources
115
ANSWER KEY
5. Check that students have correctly added the
following individuals to the pedigree: Roger,
Robert, Elizabeth, Jean, and Craig.
Chapter 4 Project Worksheet 2
(p. 91)
1. The genotypes students choose for Irene and
Leo may vary. One possible choice is for both
Irene and Leo to be heterozygous (Aa).
2. Answers will vary depending on the genotypes
selected in Step 1. For the example mentioned
above, the cells of the Punnett square should
be filled in with AA, Aa, Aa, and aa.
3. Genotypes will vary. They can be any combination of AA, Aa, and/or aa.
4. Genotypes will vary. They can be any combination of AA, Aa, and/or aa.
5. Genotypes will vary depending on the genotypes selected for the second generation and
their spouses. Check that students have completed the Punnett squares correctly and
selected genotypes for the grandchildren that
are possible given the respective genotypes of
their parents.
Section 4-1 Enrich
(p. 96)
1. Alice and Beatrice
116
Unit 1 Resources
Section 4-2 Review and Reinforce
(p. 99)
1. Cystic fibrosis
2. Recessive
3. Codominant
4. Abnormal hemoglobin
5. Hemophilia
6. Drugs to prevent infections or physical
therapy to break up mucus in the lungs
7. People with sickle-cell trait have just one
sickle-cell allele. They produce both normal
and abnormal hemoglobin but do not have
symptoms of sickle-cell disease. People with
sickle-cell disease have two sickle-cell alleles.
They produce only abnormal hemoglobin and
have the symptoms of the disease.
8. Hemophilia is more common in males than in
females because it is a recessive sex-linked trait.
9. Down syndrome is caused by an extra copy of
chromosome number 21.
10. true
11. Amniocentesis
12. karyotype
Section 4-2 Enrich
(p. 100)
1. In tropical parts of Africa, South America, and
Asia
2. In tropical Africa
3. The most likely hypothesis is that the abnormal hemoglobin in the blood of people with
sickle-cell trait makes them poor hosts for the
malaria parasite.
4. If malaria were eradicated, the sickle-cell allele
might become less frequent because people
with sickle-cell trait no longer would have an
advantage.
Science Explorer Focus on Life Science
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Section 4-1 Review and Reinforce
(p. 95)
1. Dd
2. dd
3. Dd
4. dd
5. The father has dimples.
6. 50 percent
7. X CX c
8. X CY
9. X cX c
10. X cY
11. The father is colorblind.
12. 50 percent
13. 50 percent
14. multiple alleles
15. Sex-linked genes
16. carrier
17. pedigree
2. Alice
3. Beatrice
4. Maria Cristina’s status is no longer uncertain
because she can now be assumed to have been
a carrier of hemophilia.
5. Because the trait is recessive, and females
would need to inherit two hemophilia alleles
to have the disorder
ANSWER KEY
© Prentice-Hall, Inc.
Section 4-3 Review and Reinforce
(p. 103)
1. Inbreeding involves crossing two individuals
with identical or similar sets of alleles, and offspring have alleles that are very similar to
those of their parents. Hybridization involves
crossing two genetically different individuals
so the offspring will have the best traits of
both parents.
2. Cloning is a means to produce offspring that
are genetically identical to the organism from
which they were produced.
3. Scientists splice a human gene into the plasmid of a bacterium. As a result, the bacterium
produces the protein the gene codes for.
4. Scientists insert working copies of a gene
directly into the cells of a person with a genetic disorder so the cells can produce the protein
the person is lacking.
5. It is a project to identify the DNA sequence of
every gene in the human genome so scientists
may gain a better understanding of how the
body works and what causes things to go wrong.
6. d
7. a
8. e
9. b
10. f
11. c
12. g
Chapter 4 Real-World Lab
(pp. 105–107)
For answers, see Teacher’s Edition, p. 115.
Chapter 4 Real-World Lab
(pp. 108–109)
For answers, see Teacher’s Edition, p. 124.
Section 4-3 Enrich
(p. 104)
1. CFTR prevents mucus from building up in the
lungs.
2. The viral DNA must enter the cell’s nucleus
because the first phase of protein synthesis—
the copying of the DNA into a strand of messenger RNA—takes place in the nucleus.
3. CFTR is produced on the ribosomes in the
cell’s cytoplasm.
4. Gene therapy does not lead to a permanent
cure because the genetically engineered viral
DNA dies off when the affected lung cells die.
Thus, the engineered viruses do not provide a
constant supply of CFTR.
Science Explorer Focus on Life Science
Unit 1 Resources
117