Study Guide Chapter 15 -Chromosomal Basis of Inheritance-Answers.docx

Study Guide Chapter 15 -Chromosomal Basis of Inheritance-
Answers
15.1 Mendelian Inheritance and chromosome
theory______________________________
1. Thomas Hunt Morgan identified the first solid evidence
associating a specific___gene_____ on a specific chromosome.
2. Why did Morgan choose Drosophila as his experimental
organism? (List 3 reasons)
They reproduce quickly, a new generation of adults forms every
two weeks!
Prolific (single matting = hundreds of offspring)
Only 4 chromosome pairs = simple genetics
3A. The normal phenotype for a character (phenotype most
common in nature) is called___wild type_____.
3B. Provide at least two examples this phenotype in Drosophila.
Red eyes
gray body
4A. An alternative phenotype for a character (phenotype not
common in nature) is called____mutant____.
4B. Provide at least two examples of this phenotype in
Drosophila.
White eyes
Black body
5. Morgan and his students invented a notation to symbolize
alleles in Drosophila that differed from the notation Mendel
used to represent alleles. Describe how Morgan’s Drosophila
characters are named and the symbol used for the allele type.

You can think of Morgan’s wild-type allele as equivalent to the
dominant allele in Mendel’s naming system. And you can think
of Morgan’s mutant allele as equivalent to Mendel’s recessive
allele. So whenever you have one wild-type allele and one
mutant allele, that organism will have the wild-type phenotype.
Mendel
Used the first letter of the dominant character name to represent
the dominant allele. This letter was capitalized for the dominant
allele, lower case for the recessive allele
Ex: purple flower allele dominant to white flower allele: P=
dominant allele, p=recessive allele
Morgan
-Used the first letter of the mutant character name to represent
the wild-type allele. The letter for the wild-type and mutant
alleles are both lower case. The wild-type allele gets a + sign
and the mutant does not.
Ex: red eye allele wild-type to white eye mutant: w+ = wild-
type allele, w= recessive allele.
6. Morgan and his students invented a notation to symbolize
alleles in Drosophila. Which of the following genotypes would
produce a fly that is wild-type for eye color (red vs. white
eyes)?
w+ w+
w+ w
w w
w+w+ and w+w
7. Morgan performed an experiment that yielded a 3:1 ratio of
offspring in the F2 generation; however, only the males of this

F2 generation had white eyes.
White- eye females DO EXIST, so why were only white-eye
males observed in this cross?
Complete the following chart and describe why Morgan
observed these results and how it allowed him to conclusively
determine that the gene for eye color was located on the X
chromosome.
Morgan’s experiment (see Figure 15.4) Use notation developed
by Morgan and his students
If the gene for eye color is located on the X chromosome….
P generation
Red eye Female genotype for eye color = w+w+
White eye Male genotype for eye color= w
F1 generation
Genotype of offspring =
Red eye females w+w
Red eye males w+
F2 generation (F1 M crossed to F1 F)
Genotypes of offspring:
Red eye females w+w+
Red eye females w+w
Red eye males w+
8. If the eye-color gene locus in Drosophila was located on an
autosome what would you expect to observe in the F2 offspring
produced by the crosses in the Morgan’s experiment described
in the previous question?
He would have observed equal numbers of white -eye males and
females.

15.2 Sex-linked genes inheritance
patterns______________________________________
9. Compare the X an Y chromosomes in Mammals.
X = larger, has more genes, two found in females, one in males
Y = smaller, has fewer genes, one in males, none in females
10A. Following Meiosis in mammals, each ovum (egg) contains
one __X_____(X or Y) sex chromosome, whereas following
meiosis in males each sperm cell has either an X OR Y
chromosome.
10B. Which gender M/F determines the sex of offspring in
mammals? Male
11. Your friend Bill has a “sex-linked” disease. Which
chromosome, X or Y, is more likely to contain the mutant allele
that is responsible for Bill’s disease Why?
X-because it is bigger than the Y, it has many more genes. For
this reason, if a person has a “sex-linked disease” it is most
likely on the X chromosome. It is so rare to have a sex-linked
disease as a result of a mutant gene on the Y chromosome that
sex-linked diseases are also called X-linked diseases.
12. In mammals, sex-linked genes are passed from Father’s to
__daughters__(daughters/son) only. Mother’s can pass sex-
linked genes to sons OR daughters.
13A. Name at least two sex-linked disorders found in humans.
Colorblindness, hemophilia
13B. Why are sex-linked disorders more common in males?
More common in males because males only have one X
chromosome, whereas females have two X chromosomes. If a
male inherits an X chromosome from his mom with a diseased
allele, he will have the disease because he has only one X
chromosome and does not have another X chromosome to
inherit a good (non-diseased allele). Females, because they have
two X chromosomes, have two chances to inherit a non-diseased
allele.

14. Define the term “Hemizygous”. When is this term used?
How does it differ from the term “carrier”?
The term “hemizygous” is used to describe a male who has a
mutant allele for a gene on his X chromosome. Being a male, he
does not have another X chromosome. Therefore, a male
CANNOT be a “carrier” for a sex-linked disease…he either has
the disease or he does not have the disease.
15. If a female homozygous dominant for the normal form of a
sex-linked gene mates with a male who is hemizygous for the
mutant form of the gene…
MAKE A PUNNETT SQUARE FOR THIS CROSS!
A. What fraction of the total offspring from this mating will
have the disease?
0/4 = 0 (When asked for total offspring, look at the ENTIRE
punnett square...0 out of the 4 offspring will have the disease.)
B. What fraction of the total offspring from this mating will be
carriers for the disease? 2/4 = 1/2 (When asked for total
offspring, look at the ENTIRE punnett square...2 out of the 4
offspring will be carriers.)
C. What fraction of the female offspring from this mating will
be carriers for the disease? 2/2 =1, (When asked about the
female offspring, look at the females ONLY…All females will
be carriers.)
D. What percentage of the female offspring from this mating
will have the disease?
0/2 = 0 (look at the females ONLY…0 out of the 2 females will
have the disease)
E. What percentage of the male offspring from this mating will
be carriers for the disease? NONE, Males CANNOT be
“carriers” for a sex-linked disease!
F. What percentage of the male offspring from this mating will
have the disease?
0/2 = 0 (When asked about the male offspring, look at the males
ONLY…0 out of the 2 males will have the disease.)
16. If a female heterozygous for a sex-linked disease mates with
a male who is hemizygousfor the mutant form of the gene…

MAKE A PUNNETT SQUARE FOR THIS CROSS!
A. What fraction of the total offspring from this mating will
have the disease?
2/4 = 1/2 (When asked for total offspring, look at the ENTIRE
punnett square...2 out of the 4 offspring will have the disease.)
B. What fraction of the total offspring from this mating will be
carriers for the disease? 1/4 (When asked for total offspring,
look at the ENTIRE punnett square...1 out of the 4 offspring
will be a carrier.)
C. What fraction of the female offspring from this mating will
be carriers for the disease? 1/2 (When asked about the female
offspring, look at the females ONLY…one out of the two
females will be a carrier.)
D. What percentage of the female offspring from this mating
will have the disease?
1/2 (look at the females ONLY…1 out of the 2 females will
have the disease.)
E. What percentage of the male offspring from this mating will
be carriers for the disease? NONE, Males CANNOT be
“carriers” for a sex-linked disease!
F. What percentage of the male offspring from this mating will
have the disease?
1/2 (When asked about the male offspring, look at the males
ONLY…1 out of the 2 males will have the disease.)
17. During embryogenesis, one of the X chromosomes in every
female cell becomes inactivated. What is the name given to this
inactivated X chromosome? Barr body
18A. Random inactivation of one of the X chromosomes (X
inactivation) occurs in every body cell of female mammals. In a
particular breed of cat (tortoiseshell cat) the gene for fur color
is located on the X chromosome. Two alleles exist, orange fur
allele and black fur allele. A female with a heterozyougous
genotype will be tortoiseshell (orange and black). Male
tortoiseshell cats do not exist. Explain this finding.
The heterozygous fur color genotype female cat has one X with

the orange fur allele, other X has black fur color allele. X with
orange allele is active in some cells, inactive (Barr body) in
other cells. X with black
patches of orange and patches of black fur.
A male has only ONE X chromosome, so he can only have the
orange fur color allele on his X OR the black fur color allele on
his X. He can only be orange OR black.
B. A Female orange (tortoiseshell species) cat mates with a
black male (tortoiseshell species) cat.
Orange allele= b
Black allele =B
What are the fur color genotypes of the parent cats
(REMEMBER…this allele is on the X chromosome!) ?
Female: Xb Xb
Male: XB Y
C. What are the possible genotypes of their offspring?
(construct a punnett square)
XB Xb
Xb Y
D. What are the offspring phenotypes associated with the
genotypes listed in the previous question?
XB Xb tortoiseshell female
Xb Y orange male
disorders
19. Describe how nondisjunction can occurs during meiosis.
Meiosis I = homologous chromosomes do not separate
Meiosis II = sister chromatids do not separate
20. If a gamete with an abnormal number of chromosomes
(produced as a result of nondisjunction) unites with a normal

gamete at fertilization, what condition will result? Aneuploidy
(Ex: trisomic, monosomic)
21A. A chromosome present in triplicate in fertilized egg (2n +
1 chromosomes) is said to be
_____trisomic_________________.
21B. What disease is caused by the presence of chromosome 21
in triplicate? Down syndrome
22. Breakage of a chromosome can lead to 4 types of changes in
chromosome structure. List these 4 possible changes and
describe how they occur.
Deletion –when nonsister chromatids cross over during meiosis
at wrong place
Duplication- when nonsister chromatids cross over during
meiosis at wrong place
Inversion- when when nonsister chromatids cross over during
meiosis at the right place, but one of the broken chromatids
attaches to the nonsister chromatid upside down
Translocation- when when nonsister chromatids cross over
during meiosis at the right place, but attaches to a non
homologous chromosome
Copyright © 2005 Pearson Education, Inc. publishing as
Benjamin Cummings
PowerPoint Lectures for
Biology, Seventh Edition
Neil Campbell and Jane Reece
Lectures by Chris Romero
Chapter 15
The Chromosomal Basis of Inheritance

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Concept 15.1: Where are genes located?
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Morgan’s Experimental Evidence: Scientific InquiryThomas
Hunt MorganChromosomes are the location of Mendel’s
“heritable factors”
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Morgan’s Choice of Experimental OrganismMorgan worked
with DrosophilaBreed at a high rate New generation every 2
weeks4 pairs of chromosomes
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Morgan observed:Wild type = normal phenotypes common in
the fly populations Mutant phenotypes = abnormal/non-common
Figure 15.3
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Morgan ExperimentMated:Female flies with red eyes (wild
type)Male flies with white eyes (mutant)
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Morgan ExperimentMated:Female flies with red eyes (wild
type)Male flies with white eyes (mutant)
What do you expect in the…The F1 generation =The F2
generation =
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What chromosomes differ in M vs. F?
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Morgan determined = white-eye mutant allele must be located
on the X chromosome
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on the X chromosome
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on the X chromosome
P
F1
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on the X chromosome
P
F1
F2
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on the X chromosome
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Morgan’s Team
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Concept 15.2: Sex-linked genes exhibit unique patterns of
inheritance
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Inheritance of Sex-Linked GenesSex chromosomes have genes
unrelated to sexGene located on either sex chromosome
= sex-linked gene

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Disease caused by recessive alleles on X chromosome Color
blindnessMuscular dystrophyHemophilia
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Hemophilia
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Sex-linked genesFollow specific patterns of inheritance
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Sex-linked genesFollow specific patterns of inheritance
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d. What % of offspring will be color blind
g. What % of the males from this mating will be color blind?
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X inactivation in Female MammalsMammalian femalesOne of
the two X chromosomes in each cell randomly inactivated
during development Heterozygote for gene on X?
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X inactivation in Female MammalsMammalian femalesOne of
the two X chromosomes in each cell randomly inactivated
during development Heterozygote for gene on X?
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X inactivation

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X inactivation
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X inactivation
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Concept 15.4: Alterations of chromosome Number or Structure
-
disjunctionAneuploidyTrisomic, monosomicStructure (4):
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Non-disjunction

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Aneuploidy = Fertilization of gametes in which nondisjunction
occurred
5 chromosome organism
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AneuploidyZygote is trisomic3 copies of a chromosomeEx:
Down Syndrome
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AneuploidyZygote is trisomic3 copies of a chromosome
Ex: XYY syndrome
1:1,000 men

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AneuploidyZygote is trisomic
Ex: Klinefelter's Syndrome (XXY) syndrome
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AneuploidyZygote is trisomic
Ex: Klinefelter's Syndrome (XXY) syndrome
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AneuploidyZygote is monosomic1 copy of a chromosome Ex:
Turner syndrome = monosomy X

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Alterations of Chromosome StructureBreakage of a chromosome
can lead to four types of changes in chromosome
structureDeletionDuplicationInversionTranslocation
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Deletion
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Deletion and Duplication
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Duplication

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Inversion
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Translocation
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Review!Ch. 15:Morgan’s Experiment: placed gene on
chromosomeSex-linked inheritance patternsX Inactivation in
female mammalsAlterations in chromosome number or structure
Study Guide Chapter 14-Answers
Mendel and the Gene Idea
14.1 Mendel’s 2 laws of
inheritance___________________________________________
1. Why did Mendel use pea plant as a model organism for his
study of genetic inheritance?

Peas are available in many varieties, and he could strictly
control which plants mated with which. Many of the pea plant
traits (flower color, height, etc.) varied in an “either-or”
manner.
2. Mendel tracked only those characters of the pea plants that
varied in an “either-or” manner. What does this mean? For
example, the pea plants he studied had purple OR white flowers;
there was no phenotype intermediate (light purple) between
these two varieties.
3A. Alleles: Alternative versions of genes are called alleles. For
each character, an organism inherits __2__(number) alleles for
each trait, __1___(number) from each parent.
3B. What is a dominant allele vs. a recessive allele? Give an
example.
P vs p purple flower color is dominate to white flower color
allele
3C. If an organism has one of each allele type (one dominant
and one recessive) then the __dominant__(dominant/recessive)
allele determines organism’s appearance
4. Genotypes: Homozygous dominant, Homozygous recessive
and Heterozygous Provide at least one example of each
genotype.
Homozygous dominant (Two dominant alleles) = AA
Heterozygous recessive (Two recessive alleles) = aa
Heterozygous (One dominant, One recessive allele)= Aa
5. Phenotype vs. genotype
Long fur (L) is dominant to short fur (l). Provide an example of
the three possible genotypes and the phenotype of each
genotype.

Homozygous dominant Genotype = LL, Phenotype = Long fur
Heterozygous recessive Genotype = ll, Phenotype = short fur
Heterozygous Genotype –Ll, Phenotype – Long fur
6. Describe the Typical Mendel experiment. Use the following
terms:
True-breeding
Parental (P) generation
Hybridization
Hybrids (Heterozygotes)
F1 generation
F2 generation
3:1 phenotype ratio
-Mendel always started with two homozygous (one homozygous
dominant one homozygous recessive) true-breeding organisms
in the P generation.
-He crossed P generation true-breeders (hybridization) to get F1
generation, Hybrid offspring.
-Mendel crossed plants from F1 to produce F2 generation
offspring = 3:1 phenotype offspring ratio observed
7. Construct punnett squares for the following crosses and
determine the probability of offspring genotypes and
phenotypes.
A. Rr X Rr
Genotypes: ¼ RR, ½ Rr, ¼ rr
Phenotypes: RR= dominant, Rr = dominant, rr = recessive
¾ dominant, ¼ recessive

B. AA X Aa
Genotypes: ½ AA, ½ Aa
Phenotypes: AA dominant, Aa dominant
ALL dominant
C. Dd X dd
Genotypes: ½ Dd, ½ dd
Phenotypes: Dd dominant, dd recessive
½ dominant, ½ recessive
8. What is a testcross and when is this method utilized?
The breeding of a homozygous recessive organism (aa) with an
organism of dominate phenotype but unknown genotype (AA?
or Aa?)
EX: Used when you don’t know the genotype of a dominant
phenotype organism (AA or Aa? = cross it with an organism of
KNOWN genotype = homozygous recessive organism, aa)
CROSS 1 Aa X aa
Offspring = ½ aa and ½ Aa
CROSS 2 AA X aa
Offspring = ALL Aa
So…
if you do the cross and get ½ dominant phenotype offspring and

½ recessive phenotype offspring, then you know the dominant
parent MUST have a Heterozygous genotype.
if you do the cross and get ALL dominant phenotype offspring,
then you know the dominant parent MUST BE Homozygous
dominant.
9. Explain the difference between a monohybrid and a dihybrid.
Monohybrid = follow one character P( F1 will be hybrids for
one trait (have one dominant allele, one recessive allele) (EX:
Pp)
Dihybrid = follow two characters P and R( F1 will have one
dominant allele, one recessive allele for each character (EX:
PpRr)
14.2 Laws of probability and Mendelian
inheritance______________________________
10. Multi-hybrid Cross Calculation
3 characters = trihybrid cross
Parent 1: Purple flowers (PP), Yellow (Yy), Wrinkled (rr)
Parent 2: Purple flowers (Pp), Green (yy), Round (Rr)
Parents: PPYyrr X PpyyRr
Question: What percentage of the offspring from this cross
would be predicted to have the following genotypes: PpyyRr
and PPyyRr (phenotype: purple flowers and green and round
seeds)?
Step 1. Consider each character separately (make a punnett
square for each character)
Parents: PPYyrr X PpyyRr:

PP X Pp = ½ PP, ½ Pp
Yy X yy = ½ Yy, ½ yy
rr X Rr = ½ Rr, ½ rr
Step 2. Calculate Genotype: Calculate probability for the
genotype using the Rule of Multiplication
State the Rule of multiplication:
Probability that two or more independent events will occur
together.
PpyyRr = ½ x ½ x ½ =1/8 (rule of multiplication) chance of
offspring having this genotype
PPyyRr = ½ x ½ x ½ = 1/8 (rule of multiplication) chance of
offspring having this genotype
Additional Multiplication Rule Example
Rr Female X Rr male for a character:
Eggs = ½ will have the R allele and the other ½ will have the r
allele
Sperm = ½ will have the R allele and the other ½ will have the r
allele
-Chance of RR offspring from this cross = ½ chance of having R
egg X ½ change of R sperm fusing = multiplication rule = ¼
chance!
-Chance of rr offspring from this cross = ½ chance of having r
egg X ½ change of r sperm fusing = multiplication rule = ¼
chance!
-Chance of Rr offspring from this cross = ½ chance of having r
egg X ½ change of R sperm fusing = multiplication rule = ¼
chance!

-Chance of Rr offspring from this cross = ½ chance of having R
egg X ½ change of r sperm fusing = multiplication rule = ¼
chance
Step 3. Calculate Phenotype: Use the Rule of Addition to
determine the probability of offspring that have the genotype
ppyyrr (phenotype: white flowers and green and wrinkled
seeds)?
State the rule of addition:
Probability that any one of two or more exclusive events will
occur.
PpyyRr = ½ x ½ x ½ =1/8
PPyyRr = ½ x ½ x ½ = 1/8
Add them together to get the chance of having offspring with
this PHENOTYPE
=2/8 =reduce to 1/4
Additional Addition Rule Example:
There are two ways to get a Rr offspring from the cross
described in the answer for the previous step. So we will use the
addition rule to determine the fraction of Heterozygotes
expected from this cross. Chance Rr offspring from that cross =
¼ + ¼ =1/2 (addition rule)
11. Pea plants heterozygous for pea shape and color (YyRr) are
allowed to self-pollinate. What fraction of the offspring would
be predicted to have the genotype yyrr? green and wrinkled
phenotype.
(to calculate use the method for calculating offspring
probability outlined in the previous question)
Parents= YyRr X YyRr

Step 1. monohybrid cross for each character (construct a
punnett square for each)
Yy X Yy = 1/4YY, 1/2 Yy,1/4 yy
Rr X Rr = 1/4Rr, 1/2 Rr, ¼ rr
Step 2.Multiplication rule:
yyrr= 1/4 X 1/4 = 1/16
1/16 of the offspring will be both green and wrinkled
Step 3: Does not apply to this question because there is only
one possible offspring genotype.
14.3 Inheritance patterns are more complex than predicted by
Mendel________________
12. Describe how complete dominance differs from both co-
dominance and incomplete dominance. Provide and example of
each.
Complete = Menelian characteristics where one allele is totally
dominant over another allele. Phenotypes of homozygous
dominant and heterozygous individual are indistinguishable (Ex:
PP vs. Pp)
Codominance = when both alleles for a character are dominant
(Ex: M and N molecules on RBC. MM genotype = M molecules
only, NN genotype = N molecules only, NM genotype = BOTH
N and M molecules on RBC)
Incomplete Dominance = one allele is not completely dominant
over the other allele. Resulting in a combined or mixed
phenotype in the heterozygotes.

For example, if you cross-pollinate homozygous red and
homozygous white carnation flower plants, the dominant allele
that produces the red color is not completely dominant over the
recessive allele that produces the white color. The resulting
offspring are pink.
Cross true-breeding parents in P generation (RR red and WW
white)
F1 = hybrid offspring (RW pink) because in these flowers the
red allele only contributes ½ the amount of red pigment needed
for a red flower. So to have a red flower, 2 Red alleles are
required.
F2 = take two F1 organisms (RW x RW) and cross = ¼ RR red,
¼ WW white, and ½ RW pink offspring.
13. How many alleles determine human blood type? How many
different blood type phenotypes are possible?
3 alleles: A, B and O
4 phenotypes B, A, AB, O
6 genotypes BB, BO, AA, AO AB, OO
14. Can a person with type A blood receive blood from a person
with type B blood? WHY/WHY NOT?
No, B type blood has B carbohydrates molecules on its surface.
A person with type A blood will not recognize these B
carbohydrates and their immune system will think they are
foreign invaders = blood clumps
15. A woman with type O blood has a child with type B blood.
If the “father” has Type O blood, could he be this child’s
father? NO! If the mother has type O blood, her genotype must
be OO and she can only give her child an O allele. The same is
true for a father with type O blood, he can only give the child
an O allele. The child would have to have O type blood for this

man to possibly be the father….so “He is NOT the father!”
If the “father” had type AB blood, could he have fathered this
child?
Again, Mom give the child an O allele. This man can give the
child an A allele OR a B allele. If he gave the child the B
allele, the child would have the BO genotype, giving him type B
blood. So it is possible that he could be the father of this child.
14.4 Human traits and Mendelian
inheritance___________________________________
16. Construct a pedigree following the trait for free earlobe F
(dominant) vs. attached earlobe ff (recessive) for the following
family:
Grandparents: Ed and Lucy
Children: Peggy, Sue, Dave, Martha
Peggy marries Alec
Sue marries Dan
Dave marries Lisa
Martha is single
Grandchildren:
Peggy and Alec have two boys: Jim and Bob
Sue and Dan have no children
Dave and Lisa have one girl: Betty
Ed, Peggy and Bob have attached earlobes. Everyone else has

free earlobes. Determine each person’s genotype for the earlobe
trait.
Ed (ff) and Lucy (Ff, we know she MUST be heterozygous
because she has a child (Peggy) with attached earlobes!)
Peggy (ff) and Alec (Ff): Jim (Ff), Bob (ff)
We know Alec MUST be heterozygous because he has a child
(Bob) with attached earlobes!)
Sue (Ff) and Dan (FF or Ff)
We do not know if Dan is FF or Ff because he and Sue have no
children.
Dave (Ff) and Lisa (FF or Ff): Betty (FF or Ff)
We do not know if Lisa is FF or Ff because their child also has
the dominant phenotype so her genotype is unknown.
Martha (Ff)
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Chapter 14
Mendel and the Gene Idea
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Gregor Mendel

experiments with garden peas
Figure 14.1
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Mendel’s Experimental ApproachWhy peas?
= available in many varieties
= could strictly control mating
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Mendel’s Experimental Approach
Stamens (Male)
Carpel (Female)
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Genetics Vocabulary
Alternative versions of genes = Alleles
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Organism inherits 2 alleles:
1 from mom, 1 from dad
A genetic locus is represented twice
Genetics Vocabulary
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Genetics VocabularyIf the two alleles at a locus differ…
Dominant allele = determines appearanceRecessive allele = no
noticeable effect on appearance
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Genetic Vocabulary: Homozygous vs.

HeterozygousHomozygous for a particular gene
Identical pair of alleles for that gene
Ex: PP (2 purple flower alleles)
True-breeding
- Homozygous dominant (PP)
- Homozygous recessive (pp)
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Homozygous vs. HeterozygousHomozygous for a particular
gene
Identical pair of alleles for that gene
Ex: PP (2 purple flower alleles)
True-breedingHeterozygous for a particular gene
Has a pair of alleles that are different for that gene
Ex: Pp (1 purple allele, 1 white allele)
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Homozygous or Heterozygous?
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Genetics VocabularyAn organism’s genotype (EX: Pp, PP, pp)
genetic makeupAn organism’s phenotype (Ex: Purple or white)
physical appearance

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Phenotype versus genotype
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Phenotype versus genotype
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Mendel used :Characters that varied in an “either-or”
mannerVarieties that were “true-breeding”
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Typical Mendelian Experiment
Parental Generation
Hybridization
F1 Generation
F1 self-pollinate
F2 generation

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All Purple
Hybrids
3:1
Purple : White
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Does Mendel’s segregation model account for the 3:1 ratio
observed in the F2 generation?
We can answer this question using a Punnett square
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Other pea plant characters
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Construct a Punnett Square for the following crosses:Seed
color: Y = Yellow, y = green
YY X Yy
Expected ratio observed in offspring?Seed shape: R = Round, r
= wrinkled
Rr X rr
Expected ratio observed in offspring?
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The TestcrossIn pea plants with purple flowers
Genotype is not obvious (Pp or PP)?
= Perform testcross

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The testcross
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The testcross
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Monohybrid Cross
Mendel Followed a single trait (ex: flower color)The P = true-
breeding (PP or pp)The F1 offspring = monohybrids
(heterozygous for one character) (Pp)
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Dihybrid CrossMendel followed 2 characters at the same timeP
generation = Cross two, true-breeding parents differing in two
characters
YYRR X yyrr

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Character 1
Y =YELLOW
y =green
Character 2
R=ROUND
r = wrinkled
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Mendel followed 2 characters at the same timeP generation =
Cross two, true-breeding parents differing in two characters
YYRR X yyrrF1 generation = Produces dihybrids
(heterozygous for both characters)
YyRr
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2. Independent Assortment of Chromosomes

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Homologous orient randomly at metaphase I of meiosis
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How are two characters transmitted from parents to offspring?
1. As a package? (Ex: yellow and round YR)
=Dependent Assortment
2. Independently?
=Independent Assortment
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A dihybrid cross
Only YR and yr as inherited from P generation?
YR Yr yR yr ?
Make a punnett square for each case

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Concept 14.2: The rules of probability govern Mendelian
inheritanceMultiplication RuleAddition Rule
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The Multiplication and Addition Rules Applied to Monohybrid
CrossesThe multiplication rule
Probability that two or more independent events will occur
together
Ex: coin toss
Heads ½ X Heads ½ = ¼
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Ex: Probability in a monohybrid cross
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Rule of AdditionProbability that any one of two or more
exclusive events will occur
Ex: Heterozygotes:

¼Rr + ¼rR = ½
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A multi-character cross
= two or more independent monohybrid crosses occurring
simultaneouslyCalculate the chances for various genotypes:
1. Consider each character separately
2. Go back to question being asked
3. Multiply individual probabilities together
4. Use Rule of addition (if necessary)
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Purple flowers (Pp), Yellow (Yy), Round (Rr)
Purple flowers (Pp), green (yy), wrinkled (rr)
PpYyRr X Ppyyrr
would be predicted to have purple flowers and green and
wrinkled seeds?
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1. Consider each character separately (make a punnett square
for each character)
PpYyRr X Ppyyrr:
Pp X Pp =
Yy X yy =

Rr X rr =
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for each character)
PpYyRr X Ppyyrr
Pp X Pp = ¼ PP, ½ pP, ¼ pp
Rr X rr = ½ Rr, ½ rr
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2. Go back to the original Question
PpYyRr X Ppyyrr
wrinkled seeds?
this
condition:
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PpYyRr X Ppyyrr
Pp X Pp = ¼ PP, ½ Pp, ¼ pp
wrinkled seeds?
this
condition:
Ppyyrr, PPyyrr
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3. Calculate probability for each genotype
Ppyyrr ½ X ½ X ½ = 2/16Ppyyrr
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Ppyyrr ½ X ½ X ½ = 2/16PPyyrr ¼ X ½ X ½
=1/16

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4. Rule of addition
2/16 Ppyyrr
+1/16 Ppyyrr
3/16
= chance that the offspring from this cross would have purple
flowers and green and wrinkled seeds
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A multi-character cross #2
= two or more independent monohybrid crosses occurring
simultaneouslyCalculate the chances for various genotypes:
1. Consider each character separately
2. Go back to question being asked
3. Multiply individual probabilities together
4. Use Rule of addition
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3 characters = trihybrid cross #2
white flowers (pp), Yellow (Yy), wrinkled (rr)
Purple flowers (Pp), green (yy), Round (Rr)
ppYyrr X PpyyRr
would be predicted to have white flowers and green and
wrinkled seeds?

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for each character)
ppYyRr X Ppyyrr:
pp X Pp = ½ Pp, ½ pp
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ppYyRr X Ppyyrr:
wrinkled seeds?
this
condition:
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ppYyRr X Ppyyrr:

wrinkled seeds?
otypes that fulfill this
condition:
ppyyrr
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ppYyRr X Ppyyrr:
ppyyrr ½ pp X ½ yy X ½ rr = 1/8
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Concept 14.3Inheritance patterns are often more complex than
predicted by simple Mendelian geneticsThe relationship
between genotype (Ex: Pp) and phenotype (Ex: purple) is rarely
simple
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The Spectrum of Dominance Complete dominance
Phenotypes of the heterozygote and dominant homozygote are
identical
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Codominance
Two dominant alleles affect the phenotype in separate,
distinguishable waysEx: human blood group MN
MM = RBC with M molecules
NN = RBC with N molecules
MN = ?
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Incomplete dominance
F1 hybrid phenotype is between the phenotypes of the two
parental varieties
Figure 14.10
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Dominance and Phenotype
Dominant and recessive alleles
Do not “interact”
Different alleles = synthesis of different proteins that produce a
phenotype
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Dominance and Phenotype
Dominant and recessive alleles
Do not “interact”
Different alleles = synthesis of different proteins that produce a
phenotype
Ex: flower color
White (W) vs. Red (R)
W= protein that produces white pigment
R = protein that produces red pigment
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Multiple AllelesMost genes exist in populations
In more than two allelic forms
1
2

3
1
2
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The ABO blood group in humans
Is determined by multiple alleles:
3 different alleles for enzyme I
IA = attaches the A carbohydrate
IB = attaches the B carbohydrate
i = attaches neither A nor B
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Table 14.2
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Complex inheritance patterns
Codominance
Incomplete dominance
Multiple alleles
Mendel’s fundamental laws still apply!

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Concept 14.4:Human traits follow Mendelian patterns of
inheritanceHumans = not convenient subjects for genetic
research
How can we study Human Genetics?
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Concept 14.4:Human traits follow Mendelian patterns of
inheritanceHumans = not convenient subjects for genetic
research
How can we study Human Genetics?
= Pedigree analysis
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Male =
Female =
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Follow Attached earlobe = ff

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Carriers?Disease condition = aaNo disease symptoms = Aa or
AA
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Mating of Close RelativesMating between relatives
Can increase the probability of the appearance of a genetic
disease
Cc
CC
Cc
Cc
cc

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Albinism- recessive phenotypeOnly aa
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Albinism- recessive phenotype
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Human achondroplasia phenotypeThe phenotype is determined
by a dominant allele = AA or Aa
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Human achondroplasia: Dominant allele disease
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PHENYLKETONURIA - [PKU] pp Autosomal recessive
disorderGene for phenylalanine hydroxylase (PAH), found on
chromosome 12 mutated PAH converts the amino acid
phenylalanine to tyrosine No PAH = concentration of
phenylalanine in the body can build up to toxic levels

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PKU: Recessive disease (pp)
Pp
pp
Pp
Pp
Pp
Pp
Pp
Pp/PP
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Review!Mendel’s Pea experiments:Experimental methodTypical
Mendelian experiment:
P, F1, F2Monohybrid cross vs. Dihybrid crossLaw of
Segregation and Law of Independent assortment
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Review!Solving Multi-hybrid crosses with probabilityMore
complex inheritance patterns:
Co-dominance, Incomplete dominance, Multiple allelesPedigree

Analysis
Chapter 13 Study Guide Meiosis-ANSWERS
13.1 Chromosomes are
inherited_____________________________________________
1. Chromosomes are made of segments of DNA called
__GENES_____, which are found at specific places along the
chromosome called a___LOCUS________.
2. Compare and contrast asexual and sexual reproduction.
Asexual- parent produces identical cell by mitosis = offspring a
clone of parent
Sexual- egg or sperm produced by meiosis = offspring are not
clones of parents because they have a set of chromosomes from
each parent.
13.2 Meiosis and Fertilization alternate in sexual life
cycles________________________
3. What cells undergo Meiosis? Where are these cells located in
the human body?
Certain Diploid cells located in the male or female gonad
ONLY. These cells undergo meiosis to produce gametes or sex
cells, egg or sperm. NO OTHER CELLS IN YOUR BODY
UNDERGO THIS PROCESS!
4. A human has 46 chromosomes. How many of these
chromosomes were inherited from Mom? 23
5. What is a karyotype and at what stage in the cell cycle can
one be constructed?
Ordered representation of homologous chromosomes usually
constructed from a prophase cell. Chromosomes are ordered
from largest to smallest. The sex chromosomes (XX or XY) are
placed last, at the bottom right of the karyotype.
6. What are Homologous chromosomes? What is each
homologue composed of during mitotic phases of the cell cycle?

-2 chromosomes with same length, centromere position and
order of genes (although the type of gene may be different).
One homologue was inherited from mom, the other homologue
was inherited from dad.
-Each homologue is composed of 2 sister chromatids (when the
cell is preparing to divide) until anaphase of mitosis or
anaphase 2 of meiosis.
7. What is the difference between autosomes and sex
chromosomes?
Autosome- any homologue that is not an X or Y chromosome.
All autosomes have a homologue.
Sex chromosomes- X or Y chromosome that determines gender
of organism. If you are a female XX, you have a homologous
chromosome. If you are a male XY, you do not have a
homologus chromosome for this chromosome. The X
chromosome is very large and contains many more genes
compared to the Y chromosome. We will discuss this further in
chapter 15.
8. An organism has a diploid number of chromosomes 2n = 48.
What is the haploid chromosome number? 24
13.3 Meiosis reduces chromosome
number_____________________________________
9. Describe the major event(s) that occur in each stage of
Meiosis:
Interphase- cell grows, organelles produced, copy DNA
Prophase I- chromosomes condense, homologous chromosomes
come together to form tetrads, non-sister chromatids crossover
(exchange genetic material) places where these crossing over
events occur are called chiasmata
Metaphase I- Homologous chromosomes line up together (in
tetrads) on the metaphase plate.

Anaphase I- Homologous chromosomes move to opposite poles
of the cell
Telophase I and cytokinesis- 2 haploid daughter nuclei form
followed by division of cytoplasm (cytokinesis)
Prophase II- in each daughter cell, spindle forms,
Metaphase II- sister chromatids line up on metaphase plate
Anaphase II- sister chromatids move to opposite poles of the
cell
Telophase II and cytokinesis- 4 haploid nuclei form followed by
cytokinesis to form 4 haploid cells
10. How many cell division events take place during meiosis? 2
11A. How is metaphase of Meiosis I different from metaphase
in mitosis?
Metaphase Meiosis = homologous chromosomes line up together
on metaphase plate.
Metaphase Mitosis = homologous chromosomes do not line up
together on metaphase plate.
11B. Identify and describe at least 2 more events that
distinguish meiosis from mitosis.
1. Homologous chromosomes line up together on metaphase
plate during meiosis I
2. Non-sister chromatids of homologous chromosome pairs
cross over during prophase of meiosis I
3. During anaphase I of meiosis homologues separate, during
anaphase I of Mitosis sister chromatids separate
13.4 Genetic

Variation_____________________________________________
________
12. The genetic variation observed in offspring produced by
sexual reproduction can be accounted for by what three main
behaviors of chromosomes during meiosis
1. Crossing over between non-sister chromatids of homologous
chromosome pairs during Prophase I of meiosis.
2. Independent assortment of homologous pairs on the
metaphase plate of metaphase I and II of meiosis
3. Random fertilization
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Chapter 13
Meiosis and Sexual
Life Cycles
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Asexual vs. Sexual Reproduction Asexual reproduction
one genetically identical offspring
Parent
Bud

0.5 mm
Figure 13.2
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Asexual vs. Sexual ReproductionSexual reproduction
Figure 13.1
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Concept 13.1:
Genome
Chromosomes
DNA molecule and proteins (Chromatin)
Genes
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MEIOSIS
Somatic (body) cells
Chromosomes are matched in homologous pairs
Ex: human cells have 46 chromosomes
= 23 pairs of homologous chromosomes
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Every chromosome has a match!
= homologous pair
Figure 8.12
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Every chromosome has a match!
= homologous pair
Figure 8.12
Why do we have 2 of each chromosome?
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Somatic Cells vs. Gametes
Somatic Cells
2 sets of chromosomes
= Diploid (2n)
(n = number of chromosomes in a single set)
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Somatic Cells vs. Gametes
Somatic Cells
2 sets of chromosomes
= Diploid (2n)
(n = number of chromosomes in a single set)
Gametes (eggs and sperm)
1 set of chromosomes
= Haploid (n)

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Somatic cells = diploid cell
An organism’s diploid cell has two sets of each of each
chromosome
20 from Mom; 20 from Dad
What is the diploid number of chromosomes in each of this
organism’s somatic cells?
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Somatic cells = diploid cell
An organism’s diploid cell has two sets of each of each
chromosome
20 from Mom; 20 from Dad
What is the diploid number of chromosomes in each of this
organism’s somatic cells?
2n = 40
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CHROMOSOME NUMBER AND STRUCTURE8.19 A
karyotype
ordered arrangement of a cell’s chromosomes
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Human Kayotype- 22 pairs of autosomes, 1 pair sex
chromosomes

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chromosomes condensed
A
B
C
D
-Sister chromatids
-Nonsister chromatids
-Homologous pair
-Centromere
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Somatic cells vs. Gametes
Diploid vs. Haploid
2n vs. n
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Somatic cells vs. Gametes
Diploid vs. Haploid
2n vs. n
How, when, where and why are haploid cells generated?
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Figure 8.13
How are haploid gametes produced?
= MEIOSIS!
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Overview of Meiosis- 2 cell divisions
Diploid cell
2 Haploid cells
4 Haploid cells
Separate homologues
Separate sister
chromatids

Diploid cell
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Overview of Meiosis- 2 cell divisions
Diploid 2n
2 Haploid 1n
4 Haploid 1n
Separate homologues
Separate sister
chromatids
(n = Chromosome #)
Diploid 2n
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2. Non-sisters cross over = Chiasmata
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-Tetrads line up at center of cell
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-Homologous chromosomes separate
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Meiosis
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Review: Interphase and meiosis I
Figure 13.8
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-Spindles form in each cell
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Meiosis II
- Sister chromatids line up in center of cell
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Meiosis II
- Sister chromatids separate
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Meiosis II
- 4 Haploid cells
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Telophase I, cytokinesis, and meiosis II

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Overview of Meiosis
Diploid 2n
2 Haploid 1n
4 Haploid 1n
Separate homologues
Separate sister
chromatids
n = Chromosome #
Diploid 2n
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Meiosis vs. Mitosis3 events distinguish Meiosis from Mitosis:
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Meiosis vs. Mitosis
#1. Meiosis = Homologous chromosomes pair and exchange
genetic information
- Tetrads Synapsis Crossing over (chiasmata)

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Meiosis vs. Mitosis
#2 Meiosis= Paired homologous chromosomes (tetrads) are
positioned on the metaphase plate
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Meiosis vs. Mitosis
#3 Anaphase I of Meiosis = homologous pairs move toward
opposite poles of the cell
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Meiosis vs. Mitosis
#3 Anaphase I of meiosis = homologous pairs move toward
opposite poles of the cellAnaphase II of meiosis = sister
chromatids separate
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8.15 Review: A comparison of mitosis and meiosis
Homologues pair
Homologues split
Sister Chromatids split
Sister Chromatids split
Homologues do not pair

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Concept 13.4: How does Meiosis produce Genetic Variation?
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Concept 13.4: How does Meiosis produce Genetic variation?
= Behavior of chromosomes during meiosis:
Crossing over
Independent assortment
= Fertilization:
3. Random fertilization
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1. Crossing Over - Produces recombinant chromosomes
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3. Random FertilizationThe fusion of gametes
Can produce a zygote with about 64 trillion diploid
combinations!!!
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Ch. 13Asexual vs. Sexual reproductionHomologous
ChromosomesSomatic cells= Diploid (2n)Gametes= Haploid
(1n)
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(gametes)Major events?
Prophase I, Anaphase I, Anaphase IIMeiosis vs. Mitosis?How
does Meiosis produce variation?
Chapter 12 The Cell Cycle –Study Guide ANSWERS
12.1 Cell division
basics________________________________________________
____
1. True or False: Chromosomes are made of DNA only. If false,
correct the statement False, chromosomes are made of DNA and
associated proteins. DNA is wrapped around these proteins. The
DNA is packaged in the nucleus by wrapping it around proteins.

2. Each chromosome contains many ____Genes__________,
which are segments of the chromosome that codes for a certain
polypeptide (protein).
3A. In what stages of the cell cycle are chromosomes visible
under the microscope?
The chromosomes become visible during prophase and remain
visible through mitosis (prophase( metaphase( anaphase(
telophase) and begin to de-condense during telophase of
mitosis.
B. Why are they visible during these stages?
Chromosomes duplicate and condense during prophase of
mitosis making them visible under a microscope. They condense
in order to more easily distribute the correct amount of DNA
into each daughter cell.
4. Explain the basic differences between Mitosis and Meiosis.
Name a cell type that undergoes Mitosis. Name a cell type that
undergoes Meiosis.
Mitosis = results in two identical daughter cells ex: skin cell
Meiosis = results in four non-identical haploid cells ex: cells in
gonads
5. Starting with a fertilized egg (zygote), a series of four
mitotic cell divisions would produce an early embryo with how
many cells?
1 cell ( 2 cells ( 4 cells( 8 cells( 16 cells
Each arrow represents a mitotic cell division
6A.What is the name given to the duplicated chromatids of a
chromosome when they are still attached to one another during
the first half of mitosis?

sister chromatids
6B. What are these chromatids called after they split during
anaphase of mitosis?
Chromosomes
12.2 Mitotic
phases_______________________________________________
_________
7. Specific events occur during interphase to prepare the cell for
Mitosis. Interphase is divided into 3 sub-phases. Name each
sub-phase and describe the events that occur during each sub-
phase.
G (Growth) 1= growth, manufacture organelles and proteins
S (synthesis) phase= Chromosome duplicate
G (Growth) 2 = growth, manufacture organelles and proteins
8. List (in order) starting with Interphase, the 5 phases of the
cell cycle and give a brief description of the major events that
occur in each phase. (Use the powerpoint slides as a guide for
the amount of detail you need to know about each stage)
Interphase-
Prophase
-Duplicated Chromosomes appear as 2 sister chromatids
connected at the centromere
-Mitotic spindle forms in cytoplasm from centrosomes
Metaphase

-Mitotic spindle fully formed
-Duplicated Chromosomes line up on metaphase plate located in
middle of the cell
-Each sister chromatid is attached to a spindle fiber
Anaphase
-Sister chromatids are pulled apart and taken to opposite poles
of the cell by the spindle fibers
-Cell elongates
Telophase
-Roughly opposite prophase:
-Nuclear envelopes begin to reform around chromosomes
-Chromosomes begin to uncoil
-Mitotic spindle disappears
-Mitosis complete!
9. The average human cell can complete the cell cycle in
approximately 24hrs. The cell spends most of its time in one
phase of the cell cycle. Identify this phase and explain why so
much time is spent in this phase.
Interphase- Much needs to be accomplished before a cell can
divide. The cell spends nearly the entire cell cycle in interphase
for this reason.
Interphase events:

10. Identify the function of centrosomes in mitosis.
Located outside nucleus, microtubules extend from centrosomes
to form the spindle
11. A spindle is considered complete when…? (i.e. What major
events must occur/structures form?)
Spindle microtubules are attached to sister chromatids, asters
extend from centrosomes attaching to cell membrane
12. We discussed an experiment that explained how the
chromosomes move to opposite poles of the dividing cell during
anaphase. Two hypothesis were made:
Hypothesis #1:chromosomes are pulled to the poles: spindle
fibers shorten at their spindle pole (centrosome) ends. The
spindle fiber pulls the chromosome to the spindle pole, like a
horse (chromosome) being lassoed by a rope (spindle fiber).
Hypothesis #2: chromosomes walk to the spindle poles: spindle
fibers shorten at their kinetochore (part of chromosome that is
attached to the spindle fiber) ends. The chromosome “walks
down” the spindle fiber and the spindle fiber dismantles behind
the chromosome.
A. Describe this experimental design and the experimental
results.
Experimental Design: The spindle was tagged with a fluorecent
marker. Florecence was then removed from a specific region of
the splindle at time A. (the spindle fibers in this region were
still present, only the fluorecent tag was removed). Mitosis
proceeded and another picture was taken at a later time B. The

two pictures were then compared. The region of the spindle
where the fluorescence had been removed had not moved, but
the chromosome attached to the spindle fiber had moved closer
to the spindle pole. This tells us that the spindle fiber was
shortening (dismantling) at the kinetichore (chromosome) end
and not the spindle pole (centrosome) end. If the region of the
spindle where the fluorescence had been removed had moved
closer to the spindle pole at time B compared to time A, then we
would have concluded that the spindle fiber was shortening
(dismantling) at the spindle pole (centrosome) end
B. Which hypothesis was accepted? #2
13A. Define: Cytokinesis (Hint: break this word down into
“cyto” and “kinesis”…what do these words mean?)
Cyto- as in “Cytoplasm”
Kinesis- means “ to cut”
Cytoplasm cutting!
13B. How do the processes of Mitosis and Cytokinesis differ?
Mitosis = duplicate chromosomes
Cytokinesis = divide cell into two daughter cells
13C. How does cytokinesis in animal cells and plants cells
differ?
Animal- microfilaments form ring around center of dividing cell
and contract to form two new cells
Plant- golgi vesicles containing cell wall material migrate to
center of dividing cell to form cell plate
14A. Prokaryotes reproduce by a process called ____binary
fission___________.

14B. Describe the process of prokaryote cell division using the
following terms:
origin of replication, genome replication, cell elongation, cell
division (cytokinesis)
1. Origin of replication is duplicated
2. One copy of origin of replication moves to opposite end of
cell
3. Genome Replication
4. Cell elongates
5. Cell divides
15. List at least 1 structural similarity and at least 2 structural
differences between bacterial (prokaryotic) chromosomes and
eukaryotic chromosomes
Structural Similarities:
Both are made of DNA and associated proteins
Structural Differences:
Eukaryotic chromosomes are linear, many chromosome are
present in a nucleus.
Prokaryotic chromosome circular, single chromosome NOT
surrounded by nucleus
16. List at least 1 functional similarity and at least 1 functional
difference between bacterial (prokaryotic) chromosome and

eukaryotic chromosome behavior during cell division.
Functional Similarities:
Both duplicate their DNA before the cell divides
Both divide into two cells
Functional Differences:
Eukaryotic chromosomes move to opposite poles of cell by
spindle fibers
Prokaryotic chromosome do not use spindle to move to opposite
poles of the cell
12.3 Cell Cycle
Regulation____________________________________________
______
17. Name the three checkpoints in the cell cycle. How do they
function (in general) to control the cell cycle?
G1, G2, M
The cell must have certain molecules available in certain
quantities in order to pass the checkpoint and proceed to the
next stage of the cell cycle.
18. What molecules control the G2 Checkpoint?
Cyclin dependent kinase (cdk) and cyclin. When the two
molecules join it is called MPF- Maturation Promoting Factor
19. How do cyclins and Cdk’s interact to control the G2
checkpoint?
Cdk’s are present throughout the cell cycle, whereas cyclins
accumulate just before the G2 checkpoint. Cdk’s are inactive
without cyclins. If MPF (the cdk-cyclin complex) is not present

before the G2 checkpoint, the cell cycle will not proceed. The
cell will be held in Interphase and not permitted to enter the
Mitotic phases of the cell cycle.
20A. According to the graph in the figure below, MPF activity
reaches its highest concentration at what stage in the cell cycle?
MPF concentration sharply increases just before the mitotic
phases of the cell cycle and reaches its highest point during the
mitotic phases.
20B. How does this correlate with MPF’s functions during that
stage in the cell cycle?
MPF phsophorylates (activates) proteins/molecules that have a
role in the events of mitosis. As a result, it is found in highest
concentration during this phase of the cell cycle!
Some events of Mitosis that MPF facilitates: dismantling the
nuclear membrane, building the spindle, sister chromatids
walking down the spindle fiber to the cell pole, etc.
21A. Describe how Platelet Derived Growth Factor (PDGF)
controls the cell cycle of fibro.
PDGF is released from Platelets in the area of an injury.
(Platelets are a type of blood cell present in the blood stream.
They also have a role in blood clotting.) PDGF binds to
fibroblast cells (connective tissue cells) at the site of the injury,
which causes these cells to divide. This will allow the injured
tissue to eventually heal.
21B. Is this control present in cancer cells? Why/Why not?
No. Cancer cells do not respond to normal cell cycle signals
22. Distinguish among benign, malignant and metastatic tumors.
Benign = cancer cells localized to one part of an organ or
tissue. The cells are tightly bound to one another. Due to its
structure, this type of tumor is more easily removed compared

to the other tumor types
Malignant = cancer cells that have spread throughout the tissues
of an organ. The tumor cells are no longer tightly bound to one
another. The cancer cells can and usually impair organ function.
Metastatic = cancer cells break-off from malignant tumor and
flow through blood to another area of the body. This can and
usually does lead to the formation of secondary tumors in other
parts of the body.
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Chapter 12
The Cell Cycle
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Key Roles of Cell DivisionUnicellular organisms
Reproduce by cell division
100 µm
Figure 12.2 A
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Key Roles of Cell DivisionMulticellular organisms
Development from a fertilized cell
Growth
Repair
20 µm
200 µm
(b) Growth and development.
(c) Tissue renewal. Bone marrow cells
Figure 12.2 B, C
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Concept 12.1How does Cell division results in genetically
identical daughter cells?
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Concept 12.1How does Cell division results in genetically
identical daughter cells?
Cells duplicate their genetic material before they divide
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A cell’s DNA = Genome
Genome
Chromosomes
DNA molecule and proteins (Chromatin)
Genes

50 µm
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Cell not dividing =
Chromosomes thin, loosely packed
fibers called chromatin
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Cell not dividing =
Chromosomes thin, loosely packed
fibers called chromatin
Cell dividing =
Individual Chromosomes visible
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Before a cell divides chromosomes replicate
= sister chromatids joined
together at the
centromere
Figure 8.4B

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together at the
centromere
Figure 8.4B
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together at the
centromere
Figure 8.4B
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During Cell division sister chromatids separate
2 daughter cells
each contain
a complete,

identical set of
chromosomes
Figure 8.4C
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The cell cycle
2 major phases- Interphase and Mitotic
Figure 8.5
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Interphase- 3 subphases
G1 phase
S phase
G2 phase
Figure 8.5
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Interphase
Cell parts are made (G1, G2 phase)
Chromosomes duplicate (S phase)
Mitotic phase
Duplicated chromosomes separate
Separated chromosomes are distributed into two daughter cells
What Happens During Each Phase?

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Interphase + Mitotic Phase = Cell cycle:
Interphase
Prophase
Metaphase
Anaphase
Telophase
Cytokinesis
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The 5 stages of cell division
Figure 8.6 (Part 1)
Interphase:
- Cell growth Chromosomes condense Chromosomes duplicate
Nuclear envelope begins to breakdown
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Benjamin Cummings
The stages of cell division
Figure 8.6 (Part 1)
Prophase:Duplicated Chromosomes appear as 2 sister
chromatids connected at the centromere
- Mitotic spindle forms in cytoplasm from centrosomes

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Figure 8.6 (Part 1)
Metaphase:Mitotic spindle fully formed
Duplicated Chromosomes line up on metaphase plate located in
middle of the cell
Each sister chromatid is attached to a spindle fiber
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Benjamin Cummings
Figure 8.6 (Part 1)
Anaphase:

Sister chromatids are pulled apart and taken to opposite poles of
the cell by the spindle fibers
Cell elongates
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Telophase:
Roughly opposite prophase:
Nuclear envelopes begin to reform around chromosomes
Chromosomes begin to uncoil
Mitotic spindle disappears
Mitosis complete!
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Cytokinesis:
-Division of the cytoplasm
-Animals = cleavage furrow
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Cytokinesis differs for plant and animal cells
Animals
Cytokinesis occurs
by a constriction of
the cell (cleavage)
Figure 8.7A
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Plants
membranous cell
plate (golgi vesicles) forms and splits the cell in two
Figure 8.7B

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What Phase of Mitosis?
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What Phase of Mitosis?
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8.11 Review: stages of mitosis
Figure 8.11A
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A group of cells is assayed for DNA content immediately
following mitosis and is found to have an average of 8
picograms of DNA per nucleus. Those cells would have
__________ picograms at the end of the S phase and
__________ picograms at the end of G2.
A 8 ... 8
B 8 ... 16
C 16 ... 8

D 16 ... 16
E 12 ... 16
0
*
Answer: d
Source: Barstow - Test Bank for Biology, Sixth Edition,
Question #50
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The Mitotic Spindle: A Closer LookThe mitotic spindle
Is an apparatus of microtubules that controls chromosome
movement during mitosis
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The Spindle
CentrosomesSpindle microtubulesAsters
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How are chromosomes pulled to poles?

1
1
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Benjamin Cummings
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Benjamin Cummings
?Cytokinesis usually, but not always, follows mitosis. If a cell
completed mitosis but not cytokinesis, what would be the
result?
A a cell with a single large nucleus
B a cell with high concentrations of actin
and myosin
C a cell with two abnormally small nuclei
D a cell with two nuclei
E a cell with two nuclei but with half the

amount of DNA
0
*
Answer: d
Source: Barstow - Test Bank for Biology, Sixth Edition,
Question #20
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Binary Fission: Prokaryotes (bacteria)
Origin of replication = duplicated
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Binary Fission Prokaryotes (bacteria)
One copy of origin moves to opposite end
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Binary Fission Prokaryotes (bacteria)
One copy of origin moves to opposite end
Replication continues; cell elongates
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ReviewInterphase + Mitotic phase = Cell division in
EukaryotesBinary Fission- Cell division in Prokaryotes
Similarities and Differences?Mitotic Spindle- How do sisters
move to poles?
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1
1
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Concept 12.3 How is cell cycle is regulated ?
= Molecular control system
- Controls frequency of cell division
- Ex: Skin vs. muscle cells
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The Cell Cycle Control System3 checkpoints of the cell cycle
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The Cell Cycle Control System3 checkpoints of the cell cycle
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Cell cycle stops (G0) until a go-ahead signal is received
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The Cell Cycle Control SystemG2 Checkpoint
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What are the “go-ahead” signal molecules?2 types of regulatory
proteins:
1. Cyclins
2. Cyclin-dependent kinases* (Cdks)
*activate/inactivate proteins by phosphorylation
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What are the “go-ahead” signal molecules?2 types of regulatory
proteins:
1. Cyclins
2. Cyclin-dependent kinases* (Cdks)
*activate/inactivate proteins by phosphorylation
G2 checkpoint “go-ahead” molecule
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The activity of cyclins and Cdks for G2 checkpoint
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The activity of cyclins and Cdks
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Internal and external signals
Cell cycle checkpoint is control by:
Internal signals – Ex: MPF
External signals – Ex: PDGF
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External factors for cell cycle controlGrowth factors- Stimulate
cells to dividePDGF – platelet-derived growth factor
Fibroblast cell
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Fibroblast cell
PDGF required for Fibroblast cell division

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Fibroblast cell
PDGF required for Fibroblast cell division
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PDGF Experiment
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PGDF Experiment Results

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External factors for cell cycle control
1. Anchorage dependence
2. Density-dependent inhibition
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Cancer cells
Exhibit neither density-dependent inhibition nor anchorage
dependence
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Loss of Cell Cycle Controls in Cancer CellsCancer cells
Do not respond normally to the body’s control mechanisms
Form tumors:
Benign, malignant, metastatic
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Benign, Malignant and Metastatic tumors
Figure 12.19
2

malignant
4
3
1
Benign
Tumor
Glandular
tissue
Cancer cell
Blood
vessel
Lymph
vessel
Metastatic
Tumor
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???????????????????????????????????????????List differences
between:Mitosis (eukaryotic cells)Binary Fission (prokaryotic

cells)
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?Differences between mitosis in eukaryotic cells and binary
fission in prokaryotic cells:Nuclear envelopeSpindleSeveral
chromosomes vs. one
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ReviewCell division- Reproduction (unicellular)Cell division-
Development, Growth, RepairGenetic material duplicated and
divides BEFORE cell divides. WHY?
Chromosome
Sister Chromatids
2 Chromosomes
Benjamin Cummings
ReviewInterphase + Mitotic phase = Cell division in
EukaryotesBinary Fission- Cell division in Prokaryotes
Similarities and Differences?Mitotic Spindle- How do sisters
move to poles?
Benjamin Cummings

Review!Ch. 12:Cell Cycle Control
Molecular signals
Internal
G2 = MPF (cyclin +CdK)
ExternalCancer
DATE___________________
Chapter 12- Cell Cycle
1. Phases of the cell cycle- An organism’s body cells have 4
chromosomes.
A. Identify the major characteristics of each phase.
B. OPTIONAL-Draw a picture to illustrate these characteristics.
Cell Cycle Phase
A. Characteristics of phase
B. OPTIONAL-Illustration of phase
Interphase
G1-
S-
G2-
Prophase
Metaphase
Anaphase

Telophase
Cytokinesis
-A researcher treats cells with a chemical that prevents DNA
synthesis. This treatment traps the cells in which part of the cell
cycle?
#2 OPTIONAL PRACTICE
2. Phases of the cell cycle- An organism’s body cells have 2
chromosomes
A. Identify the major characteristics of each phase.
B. Draw a picture to illustrate these characteristics.
Cell Cycle Phase
A. Characteristics of phase
B. Illustration of phase
Interphase
G1-
S-
G2-
Prophase
Metaphase
Anaphase

Telophase
Cytokinesis
3. During anaphase, do kinetochore microtubules:
Hypothesis #1: shorten at their spindle pole ends?
Hypothesis #2: shorten at their kinetochore ends?
EXPERIMENTAL RESULT:
-CONCLUSION:
-What observation would have to have been made to support the
OTHER hypothesis?
4A. Cyclin combines with Cyclin Dependent Kinase (CdK) to
form Maturation Promoting Factor (MPF). The Cyclin
concentration and MPF activity during the cell cycle are shown
in the figure below. Describe where a line on the graph would
be drawn to represent the CdK concentration through the cell
cycle.
B. Using your understanding of the molecules that control the
G2 checkpoint and the graph above, make at least one statement
about when these molecules are present & absent during the cell
cycle and how this results in cell cycle control.
5. Tumors
Benign
Malignant

Metastatic
Describe basic structure of this tumor.
Cancerous cells?
Localized to single tissue/organ?
Prognosis (good/fair/poor)
Typical treatment?
Chapter 13-Meiosis and Sexual Life Cycles
1. Important Terminology: Match the terms listed below with
the appropriate letter in the figure below.
Sister chromatids
Nonsister chromatids
Homologous pair
Centromere

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ABCD
2. Describe the differences between the somatic cell s and
gametes in your body.
Somatic cell
Gamete
Number of chromosomes
Ploidy (haploid or diploid)
Example
3. Phases of Meiosis- An organism’s body cells have 4
chromosomes (2 pairs)
A. Identify the major characteristics of each phase that differs
from Mitosis.
B. OPTIONAL-Draw a picture to illustrate these characteristics.
Meiosis Phase
A. Characteristics of phase that differs from Mitosis
B. OPTIONAL-Illustration of phase
Interphase
Prophase I
Metaphase I

Anaphase I
Telophase I & cytokinesis
Prophase II
Metaphase II
Anaphase II
Telophase II & cytokinesis
#4 OPTIONAL PRACTICE
4. Phases of Meiosis- An organism’s body cells have 2
chromosomes (1 pair)
A. Identify the major characteristics of each phase that differs
from Mitosis.
B. Draw a picture to illustrate these characteristics.
Meiosis Phase
A. Characteristics of phase that differs from Mitosis
B. Illustration of phase
Interphase
Prophase I

Metaphase I
Anaphase I
Telophase I & cytokinesis
Prophase II
Metaphase II
Anaphase II
Telophase II & cytokinesis
5. Fruit flies have a diploid number of 8, and honeybees have a
diploid number of 32. Assuming no crossing over, is the genetic
variation among offspring from the same two parents likely to
be greater in fruit flies or honeybees? Explain.
Chapter 14-Mendel and the Gene Idea
1. Genetics Terminology
Match each commonly used genetics term with its appropriate
definition or example.
TERMS:DEFINITIONS AND EXAMPLES:

__ heterozygous
a. Blue-eyed blonde mates with brown-eyed brunette
__ homozygous
b. BB or bb
__ monohybrid cross
c. not on sex chromosomes
__ autosomal
d. blue or brown eyes
__ genotype
e. Bb
___ phenotype
f. locus on a chromosome that codes for a given polypeptide
__ gene
g. Blonde mates with brunette.
__ allele
h. BB, Bb, or bb
__ dihybrid cross
i. Males have only one for each gene on the X chromosome
2. Make a punnett square using the following information.
Traits: Oval eyes = A, Round eyes = a

Parents: Mom Aa, Dad aa
-What eye shape does Mom have?
-What eye shape does Dad have?
-What fraction of the offspring will have oval eyes?
-What fraction of the offspring will have round eyes?
-What fraction of the offspring will have the Homozygous
Dominant genotype AA?
-What fraction of the offspring will have the Heterozygous
genotype Aa?
Recessive genotype aa?
3. Make a punnett square using the following information.
Traits: Brown eyes = B, Blue eyes = b
Parents: Mom Bb, Dad Bb
-What eye color does Mom have?
-What eye color does Dad have?
-What fraction of the offspring will have brown eyes?
-What fraction of the offspring will have blue eyes?
Dominant genotype BB?
-What fraction of the offspring will have the Heterozygous

genotype Bb?
Recessive genotype bb?
4. Multi-hybrid cross #1:
Parent 1: Purple flowers (Pp), Yellow (Yy), Round (Rr)
Parent 2: Purple flowers (Pp), green (yy), wrinkled (rr)
Parents: PpYyRr X Ppyyrr
would be predicted to have the following genotypes: Ppyyrr,
PPyyrr
for each character)
Parents: PpYyRr X Ppyyrr:
Pp X Pp =
Yy X yy =
Rr X rr =
2. Calculate probability for each genotype using the Rule of
Multiplication
Ppyyrr ½ x ½ x ½ = 2/16
PPyyrr
3. Use the Rule of Addition to determine the probability of
offspring that have the following genotypes:
Ppyyrr =2/16

PPyyrr =
5. Multi-hybrid Cross #2
Parent 1: White flowers (pp), Yellow (Yy), Wrinkled (rr)
Parent 2: Purple flowers (Pp), Green (yy), Round (Rr)
Parents: ppYyrr X PpyyRr
would be predicted to have the following genotypes: ppyyrr
(phenotype: white flowers and green and wrinkled seeds)?
for each character)
Parents: ppYyrr X PpyyRr:
pp X Pp =
Yy X yy =
rr X Rr =
2. Calculate probability for the genotype using the Rule of
Multiplication
ppyyrr=
3. Use the Rule of Addition to determine the probability of
offspring that have the genotype ppyyrr (phenotype: white
flowers and green and wrinkled seeds)?
6. Pedigree for a recessive trait. Determine the genotype and
phenotype of each individual in the pedigree shown below. Use
A for dominant, a for recessive.

7. Joan was born with six toes on each foot, a dominant trait
called polydactyly. Two of her five siblings and her mother, but
not her father, also have extra digits. Draw a pedigree inclucing
all family members mentioned in the question. Use D and d to
symboloze the alleles for this character. What is Joan’s
genotype for the “number-of-digits” character?
Chapter 15-The Chromosomal Basis of Inheritance
1. A heterozygous brown-eyed human female who is a carrier of
color blindness marries a blue-eyed male who is not color-blind.
Color blindness is a sex-linked trait. Assume that eye color is
an autosomal trait and that brown is dominant over blue. What
is the probability that any of the offspring produced have the
traits listed? Construct two punnett squares, one for hair color
and one for color blindness.
Eye color (autosomal trait):
B
b
b
b
Color blindness (sex-linked trait):
XA
Xa
XA
Y

a. Brown eyes
b. Blue eyes
c. Color blind OFFSPRING?
d. What fraction of the MALE OFFSPRING will be color-blind?
e. What fraction of the FEMALE OFFSPRING will be color-
blind?
f. What fraction of the FEMALE OFFSPRING will be carriers
for colorblindness?
g. What fraction of the MALE OFFSPRING will be carriers for
colorblindness?
h. What fraction of the TOTAL OFFSPRING will have Brown-
eyes and be color-blind?
i. Why do males show sex-linked traits more often than
females?
2A. Describe the process of X inactivation in female mammal
body cells.
2B. Why does this process not occur in male mammal body
cells?
2C. Discuss at least one possible reason for this phenomenon.
3. Construct a linkage map using the following gene
recombination frequencies.

The Recombination Frequency between characters:
A and B = 30%, A and C = 20%, and B and C = 10%.
4. Rip two long strips of paper from a piece of scrap paper. On
the end of each strip of paper write “A B C D”. These letters
represent gene alleles on non-sister chromosomes that are
crossing over during prophase I of meiosis. Rip one strip
between the B and C and Rip the other strip between the C and
D. Transfer the pieces you ripped off to the other non-sister.
Record the sequence of alleles on each non-sister below.
Sequence on non-sister 1:
Sequence on non-sister 2:
-What type of chromosome alterations have occurred?
PAGE
16
A
B
C
D

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