2. What are the key components of
chromosomes?
A. DNA
-heterochromatin
-euchromatin
B. Proteins
C. Found in nucleus
D. You should
understand the relationship between
DNA and proteins (chromatin packing
and histones)
3. Key terms
A. Eukaryotic chromosomes-made of DNA
and proteins (histones)
B. Gene-heritable factor that controls
specific characteristics
-made up of a length of DNA, found on a
specific chromosome location (a locus)
C. Allele-one specific form of a gene (all
found at the same locus)
-Example: Everyone has the gene for eye
color. The possible alleles are
blue, brown, green, etc.
4. More Key Terms
D. Genome-total genetic material of an
organism or species (Example: The
Human Genome)
E. Gene pool-total of all genes carried by
individuals in a population
6. Mutations
B. Gene mutation-involves changes in single base
pairs
-Some mutations may not have any effect on the
cell and may involve:
1. part of the sense strand of DNA which is
not transcribed
2. part of the DNA that a cell does not use
3. changes in second or third bases of a
codon (since the genetic code is degenerate
the same base may still be coded for)
7. Mutations
B. Gene mutation-involves changes in single base pairs
Example: Insertion or deletion of single organic bases
-changes the DNA sequence that will be transcribed
and translated
original DNA sequence: ATG-TCG-AAG-CCC
transcribed: UAC-AGC-UUC-GGG
translated: tyr-ser-phe-gly
addition of base A: ATA-GTC-GAA-GCC-C
transcribed: UAU-CAG-CUU-CGG
translated: thy-glu-leu-arg
8. Mutations: Base substitutions and
sickle-cell anemia
A. Hemoglobin-protein that helps RBC carry
oxygen
B. Hb is a gene that codes for hemoglobin
-made of 146 amino acids
C. In some cases one base is substituted for
another
normal: (HbA) base substitution: (HbS)
CTC CAC
GAG GUG
-after transcription and translation HbA produces
glutamic acid and HbS produces valine
9. Mutations: base substitutions and
sickle-cell anemia
D. The altered hemoglobin HbS is crystalline
at low oxygen levels causing the RBC to
become sickled and less efficient at
oxygen transport
E. Symptoms of sickle cell anemia
-physical weakness
-heart or/and kidney damage
-death
10. Mutations: base substitutions and
sickle-cell anemia
F. In heterozygous people (one normal allele and one
sickle cell allele)
-the alleles are codominant, but the normal allele is
expressed more strongly
-in codominance both alleles are expressed (one is
not dominant to the other)
-some sickled cells present, but most are normal
-some people show mild anemia (deficiency of the
hemoglobin, often accompanied by a reduced
number of red blood cells and causing paleness,
weakness, and breathlessness)
11. Mutations: base substitutions and
sickle-cell anemia
G. Advantages of being heterozygous
-In areas where malaria is infested:
-Plasmodium cannot live in erythrocytes with HbS
-Heterozygous individuals have a reduced
chance of contracting the protist that is carried by
mosquitoes
12.
13. Karyotyping
A. Karyotyping-process of finding the
chromosomal characteristics of a cell
-chromosomes are stained to show banding
and arranged in pairs according to size and
structure
-You should be able to
look at a karyotype
and determine the
sex of the individual
and if non-disjunction
has occurred
14. Amniocentesis and karyotyping
A. Amniocentesis
-performed at around 16 wks
-sample of amniotic fluid is taken and cultured
-when there are enough cells, a karyotype can be
performed
-chromosomes are arranged into pairs to detect
any abnormalities
-can be used to detect Down’s syndrome (a.k.a.
trisomy 21)
-can be used to recognize sex or non-disjunction
16. Karyotyping and Chorionic villus
sampling
Sampling is performed around 11 weeks of pregnancy
Cells are gathered from chorionic villi (cells from the zygote)
Cells are cultured, DNA is extracted and a karyotype is
made
18. Meiosis Key Terms
A. Diploid-having two sets of chromosomes
B. Homologous chromosomes-matching
pairs of chromosomes
-have the same genes
-are not identical (one chromosome comes
from each parent, thus alleles may be
different)
-found in diploid cells
C. Haploid-having only one set of
chromosomes
19. More Meiosis Terms
D. Chromatids -two parts of a chromosome
E. Centromere -part of a chromosome that
connects the chromatids
F. Reduction division-in organisms that reproduce
sexually
-reduction of the number of chromosomes by
half (from a diploid nucleus to a haploid nucleus)
-think of eggs and sperm; both are haploid
(when they unite the diploid number is restored)
20. Meiosis
B. Produces gametes (sperm and egg)
C. Overview
1. homologous chromosomes pair (diploid)
2. two divisions (meiosis I and meiosis II)
3. result=4 haploid cells
D. When gametes unite (to produce a diploid cell)
a cell with homologous pairs is created
-one set of chromosomes is from the mom
and one set of chromosome is from the dad
21. chiasmata
Meiosis (details) Tetrad
A. Interphase -DNA replication
B. Prophase I
-chromosomes condense
-nucleolus becomes visible
-spindle formation
-synapsis-homologous chromosomes are side
by side (they become a tetrad and are
intersected at the chiasmata)
-nuclear membrane begins to disappear
22. Meiosis (details)
C. Metaphase I: Bivalents move
to equator
D. Anaphase I:
-Homologous pairs split
-One chromosome from each
pair moves to opposite pole
E. Telophase I:
-Chromosomes arrive at poles
-Spindle disappears
23. Meiosis (details)
F. Prophase II – new spindle is formed at
right angles to the previous spindle
G. Metaphase II – Chromosomes move to
equator
H. Anaphase II –
-Chromosomes separate
-Chromatids move to opposite poles
24. Meiosis (details)
I. Telophase II
-Chromosomes arrive at poles
-Spindle disappears
-Nuclear membrane reappears
-Nucleolus becomes visible
-Chromosomes become chromatin
-Cytokinesis
25. Crossing over of homologous
chromosomes
A. An important source of variation
-creates new combinations
B. Happens during prophase I
C. Called a synapsis
D. Recombination-reassortment of genes into
different combinations from those of the parents
Chiasmata formed during a synapsis
26. Crossover
H E Start
A A genotype HE
B B genotype HE
C C genotype he
D genotype he
D
h e
H E
A
Crossover
B and C become B
recombinants C
D
h e
H E Results
A A genotype HE
B B genotype He
C C genotype hE
D genotype he
D
h e
27. Meiosis and genetic variation
A. The number of possible gametes
produced by random orientation of
chromosomes is 2n (where n is = to the
haploid number of chromosomes)
B. In humans the production of 1 gamete has
over 8 million possible combinations (223)
C. Recombination (in prophase I) + Random
orientation of chromosomes (in
metaphase I) = an infinite number of
variations
28. Meiosis and Non-disjunction
A. Disjunction - when the homologous
chromosomes separate in anaphase I
B. Aneuploidy -happens when chromosomes
do not separate (in anaphase I or II)
-caused by non-disjunction
-result: one cell missing a chromosome and one
cell having an extra chromosome
-Total number of chromosomes = 2n 1
C. Polyploidy- having more than two complete
sets of chromosomes (common in plants)
30. Non-disjunction and Down’s
syndrome
A. One of the parent gametes contains two
copies of chromosome 21
B. The zygote then ends up with 3 copies
-2 from one parent
-1 from the other
C. Down’s syndrome = trisomy 21
D. Chances of non-disjunction of
chromosomes increases with age in
females (in males too, but less of an
effect)
31. Non-disjunction and Down’s
syndrome
E. Female age has a greater effect because:
-gametes are produced before birth
-more exposure to
chemicals, radiation, etc.
F. Male age has less effect because they do
not produce gametes until puberty
G. Genetics may also increase the likelihood
of having a child with Down’s syndrome
32. Theoretical Genetics Key Terms
A. Dominant allele-the allele that always shows in
the heterozygous state (Example: Bb=brown)
B. Recessive allele-the allele that only shows in
the homozygous recessive state (Example:
bb=blue)
C. Codominant alleles-pairs of alleles where two
differing alleles are shown in the phenotype in
a heterozygote
D. Homozygous -having two identical alleles of a
gene (Example: BB or bb)
E. Heterozygous -having two different alleles of a
gene (Example: Bb)
33. More Vocabulary
F. Carrier- a person who has a recessive
allele, but does not express it (they are
generally heterozygous, Bb)
G. Genotype-alleles that a person has
(the letters) Ex: Bb
H. Phenotype- the physical characteristics
the a person shows (caused by the
genotype) Ex: brown hair or blue eyes
I. Test cross- crossing two or more
genotypes to find the possible genetic
outcomes
34. Mendel’s Monohybrid Crosses
A. Punnett square-shows possible outcomes from
a test cross
B. Mendel studied characteristics of pea plants
Wrinkled and round peas
(round peas are dominant)
35. Gregor Mendel’s Findings
Dominant Recessive
Trait Expression Expression
1. Form of ripe seed Smooth Wrinkled
2. Color of seed albumen Yellow Green
3. Color of seed coat Grey White
4. Form of ripe pods Inflated Constricted
5. Color of unripe pods Green Yellow
6. Position of flowers Axial Terminal
7. Length of stem Tall Dwarf
36. Mendel’s Monohybrid Crosses
C. Mendel found tall is dominant over short
D. His procedures were:
1. Start with 2 pure breeding homozygous
plants (This is the P generation.)
-Plant 1=tall (TT)
-Plant 2= short (tt)
2. Cross-breed the plants
-Place pollen from the tall plant in the short
plant and vice versa.
37. Mendel’s Monohybrid Crosses
D. His procedures were:
3. The F1 generation is the 1st group of
offspring.
-All were tall, and had a heterozygous
genotype. (Tt)
-This is an application of the law of
segregation.
-All offspring had a ‘T’ from one parent
and a ‘t’ from the other parent
38. Mendel’s Monohybrid Crosses
D. His procedures were:
4. The F1 offspring were then crossed. (Tt x Tt)
-Possible outcomes can be found using a
Punnett square
GENOTYPES T t
-25% TT T TT Tt
-50% Tt
-25% tt t Tt tt
PHENOTYPES
-75% Tall -25% Short
39. Multiple alleles
A. When genes have more than two alleles
B. Example: Blood type has 4 phenotypes based
on three alleles (IA, IB and i)
C. IA and IB are codominant and i is recessive
D. This is why parents can have kids with
different blood types
Phenotypes Genotypes
A IAIA or IAi
B IBIB or IBi
AB IAIB
O ii
40. Multiple alleles
D. Perform a test cross for P: mother with O blood
type and father with AB blood type. What are
the possible phenotypes?
Phenotypes Genotypes
A IAIA or IAi
B IBIB or IBi
AB IAIB
O ii
41. Multiple alleles
Perform a test cross for P: mother with O blood
type and father with AB blood type. What are the
possible phenotypes for F1?
P = ii x IAIB
i i
IA IA i IA i
IB IB i IB i
None of the children can have the same blood type as
either of the parents.
42. Codominance
A. When neither allele for a gene is recessive
B. Example: Blood type
C. Alleles A and B are both dominant (both are
expressed)
D. i is recessive to alleles A and B
E. One letter is chosen and the possible alleles
are written in upper case letters to illustrate
codominance Phenotypes Genotypes
A IAIA or IAi
B IBIB or IBi
AB IAIB
O ii
43. Sex chromosomes and gender
Only possibility for P generation = XX and XY
X X The sex of all babies
is determined by the
X XX XX chromosomes in the
sperm from the
Y XY XY man.
-The X chromosome is larger and carries more
genes than the Y chromosome
-Examples of genes on X, but not Y = color
blindness and hemophilia
-Many sex-linked traits are related to the X
chromosome.
44. Sex linkage
A. Genes carried on sex chromosomes
(usually X)
B. Example: Hemophilia-a blood disorder
that prevents clotting
-patients do not produce clotting factors that
allow coagulation of blood, and thus torn
blood vessels are prevented from closing
-most common in boys (they get it from their
mother’s X chromosome, as they only get
one X, which means only one chance to get
the dominant allele)
45. Sex linkage
C. Two parents without hemophilia:
XHXh and XHY
XH Xh
XH XHXH XHXh
Y XHY XhY
*The XhXh does is very rare
*Males cannot be heterozygous carriers because they
only have one X.
*Females can be carriers and pass on the trait to the next
generation. They can be heterozyg. or homozyg.
* XH -Normal and Xh –Hemophilia
46. Predict the genotypic and
phenotypic ratios of monohybrid
crosses for each of the following.
1. Sickle cell anemia: HbA=normal and HbS=sickle cell
HbAHbS x HbAHbS
2. Colorblindness: XB=normal and Xb=colorblind
XBXb x XbY
3. Hemophilia: XH=normal and Xh=hemophilia
XhXh x XHY
4. Blood type: IAi x IBi
**You should be able to determine if alleles are codominant
because both alleles will be represented by capital letters.
You should also know if the inherited traits are sex-linked.
47. Mendel’s Law of Segregation
A. States: The separation of the pair of
parental factors, so that one factor is
present in each gamete. (This is how it is
written in the IB book.)
B. The two alleles for each characteristic
segregate during gamete production.
This means that each gamete will contain
only one allele for each gene. This allows
the maternal and paternal alleles to be
combined in the offspring, ensuring
variation. (This is from wikipedia.)
48. Mendel’s Law of Segregation and
Meiosis
A. Mendel looked at genes (on chromosomes)
B. Found that each gene appeared twice (in
homologous pairs)
C. Figured out that when a synapsis occurs in
prophase I followed by a separation in
anaphase I, homologous chromosomes
move to opposite poles
D. One chromosome from each pair ends up
in a gamete
49. Mendel’s Law of Independent
Assortment
A. States that allele pairs separate
independently during the formation of
gametes
B. Any one pair of characteristics may
combine with any one of another pair of
characteristics
C. See p. 163 in Green Book
50. Independent assortment and
meiosis
A. Any combination of chromosomes is possible in
metaphase I (there is no ‘master plan’ for the
order that they line up on the metaphase plate
before separation)
B. Mendel thought all genes were inherited
separately and had no relationship
-Ex: Pea plants could be green or yellow and wrinkled or
round. Shape and color had nothing to do with each
other, because the genes are on separate
chromosomes. Any combination could have been
produced
(wrinkled/green, wrinkled/yellow, round/green, round
yellow)
C. This is demonstrated in Dihybrid crosses.
D. Today we realize that there are many exceptions
52. Mendel’s Law of Independent Assortment
A. Any one of a pair of characteristics may combine
with either one of another pair
B. Example:
gene=shape of pea
alleles=round (R) or wrinkled (r)
gene=color of pea
alleles = yellow (Y) or green (y)
*When crossing two plants that are heterozygous
for both traits the offspring will show all
combinations. This shows that genes for shape
and color are independent (unlinked).
53. Dihybrid crosses
Parent genotypes: SsYy x SsYy
-S=smooth s=wrinkled
-Y=Yellow y=green
SY Sy sY sy
Possible
allele
SY SSYY SSYy SsYY SsYy
combos
from one Sy SSYy SSyy SsYy Ssyy
parent
sY SsYY SsYy ssYY ssYy
sy SsYy Ssyy ssYy ssyy
54. F1 Genotypic ratios
Possible ratios for 1:SSYY
SsYy x SsYy 2:SSYy
1:SSyy
SY Sy sY sy 2:SsYY
4:SsYy
SY SSYY SSYy SsYY SsYy 2:Ssyy
1:ssYY
Sy SSYy SSyy SsYy Ssyy 2:ssYy
1:ssyy
sY SsYY SsYy ssYY ssYy F1 Phenotypic ratios
9: smooth-yellow
3:smooth-green
sy SsYy Ssyy ssYy ssyy
3: wrinkled-yellow
1: wrinkled-green
55. Perform a test cross for
P: SSYY x ssyy
1. What will the genotype and phenotype ratios be
for the F1 generation?
2. What will the phenotype and genotype ratios be
for the F2 generation?
3. Determine the recombinants in each generation.
56. Perform a test cross for P generation: SSYY x ssyy
SY SY SY SY
sy SsYy SsYy SsYy SsYy
sy SsYy SsYy SsYy SsYy
Sy SsYy SsYy SsYy SsYy
Sy SsYy SsYy SsYy SsYy
All F1 generation offspring are heterozygous (SsYy).
What will the outcome be if you cross two individuals
from F2?
57. Autosomal Gene Linkage
A. Autosome -all chromosomes that are not
sex chromosomes
B. Sex chromosomes -determine if an
individual is male or female
C. Linkage group -a group of genes whose
loci are on the same chromosome
D. Gene linkage is caused by pairs of genes
being inherited together. The presence or
absence of one can affect the other.
58. Autosomal Gene Linkage
A. In gene linkage, all of the genes on a
chromosome are inherited together
B. Does not apply to Mendel’s law of independent
assortment
C. The closer the loci of the two genes on the
chromosome, the smaller the chance that
crossing over will occur in a chiasmata
D. If no info. is given, assume the genes are not
linked
59. Autosomal Gene Linkage
E. These genes do not
follow the law of F1 generation
independent assortment. PL PL PL PL
F. Example:
pl
P=purple, p=red,
L=long, l-round pl
P gen.=PPLL x ppll
pl
pl
*all PpLl
60. Autosomal Gene Linkage
F2 generation
F. Example continued:
P=purple, p=red,
PL Pl pL pL
L=long, l-round PL
F1 gen.=PpLl x PpLl
Pl
-predicted results would be a
ratio of 9:3:3:1 pL
-observed results were very
pL
different than predicted
because the genes are for
color and pollen shape are
linked on the same
chromosome. See p. 87 in your review guide.
61. Gene linkage and Drosophila
A. When genes are linked actual outcomes do
not match expected outcome
B. This would be evident in a dihybrid cross (a
cross between two genes)
C. Linked genes are represented in vertical
pairs with horizontal lines between them
Example: b+=tan body b=black body
w+=long wings w=short wings
b+ w+ *The fly is tan and has long wings.
b+ w+ (genotype b+b+w+w+)
63. Recombinants and gene linkage
A. Recombinants are formed as a result of
crossing over (during prophase I)
B. They are found in combinations that did
not exist in the parents
64. Polygenic Inheritance
A. When the inheritance/expression of a
characteristic is controlled by more than one
gene
B. Example: Human skin color
-involves at least 3 independent genes
-A, B and C represent alleles for dark
-a, b and c represent alleles for light skin
-P: AABBCC x aabbcc
-F1: AaBbCc (all are heterozygous for all alleles)
-F2:???
65. Polygenic Inheritance
B. Example: Human skin color
-F2:???
-to find out you can make a Punnett
square
-The more dominant alleles there are, the
darker the skin
-Dominant alleles are codominant (they
are all expressed)
67. Polygenic Inheritance
Example: Flower color of beans
-Genes A and B control color expression
-AP and BP = purple
-AW and BW = white
-Each color has two alleles
-Purple and white are codominant
-Because of the codominance the flowers will be
shaded depending on their genotypes
68. P: APAPBP BP x AWAWBWBW
F1: APAWBPBW
AP BP AP BP AP BP AP BP
AWBW
AWBW
AWBW
AWBW
69. F1: APAWBPBW x APAWBPBW
F2: See Punnett square (fill it in)
Determine which ones are white, purple and
intermediates.)
AP BP AP BW AWBP AWBW
AP BP
AP BW
AWBP
AWBW
70. Polygenic inheritance and variation
patterns
Two patterns are commonly expressed
A. Continuous variation – shows a continuum of
variation among a population in a bell curve
format
- Example: In height expression people can be
short, medium or tall (and everything in
between)
B. Discontinuous variation – does not show a
continuum (it is one or the other, there is no in-
between expression of phenotype)
- Example: Blood type can be A, B, AB or O
(nothing in between)
71. Pedigree Charts
A. Used to show inheritance of traits over
several generations
B. Affected individuals-shaded black
C. Unaffected individuals- left blank
D. Heterozygous individuals (carriers)-
shaded grey or filled in half way
E. Example: Queen Victoria and
hemophilia (recessive/sex-linked)
72.
73. PCR
A. PCR=polymerase chain reaction
B. PCR=DNA photocopier
C. Used to make copies of specific DNA
sequences (for study)
74. How a PCR works
1. DNA is heated (H bonds are broken)
2. RNA primers are added to start
replication
3. As the DNA cools primers bind to the
single strands (H bonds and
complementary base pairing)
4. Nucleotides and DNA polymerase are
added
75. How PCR works
5. Nucleotides bond with ‘exposed’ bases on
the DNA strand
6. DNA polymerase joins them
7. Complimentary strands are formed
8. New strands can be heated, separated
and copied
9. Animation
76. Gel electrophoresis
A. Technique used to separate molecules
base on their rates of movement in an
electric field
-caused by charge and size of molecules
B. Commonly used in DNA profiling
77. Gel electrophoresis and DNA
profiling
A. Scientists cut a mixture of DNA into segments
by restriction enzymes
B. Place into a special gel with a current running
through it
C. DNA separates into bands according to size
D. Mixture is compared to a control group
E. The more similar the DNA strands are the
more closely they are related
F. See p. 30 in review guide or p. 66 in IB
textbook for examples
78. Figure 3.
Comparison of known gel results for normal
hemoglobin (AA), sickle cell disease (SS) and
sickle cell trait (AS). S represents the
molecular size marker. What results are
present in the lanes marked with a question
mark?
79. Application of DNA profiling
A. Criminal investigations
-collect blood or semen from suspect
-if enough is not present use PCR
-compare gel electrophoresis of suspect
and victim
B. Indentify remains of deceased
-take blood samples of living
-compare with samples from dead
C. Paternal tests
81. Genetic Screening
B. Advantages:
1. pre-natal diagnostics (seek treatment or
abortion) Ex: PKU or down’s syndrome
2. Reduce frequency of alleles that cause
genetic illnesses (opt to not reproduce or
use IVF and select embyros without genetic
defect)
3. Genetic diseases that show symptoms in
later life can be detected earlier (Ex:
Huntington’s disease)
82. Genetic Screening
C. Disadvantages:
1. Increased abortion rate
2. Harmful psychological effects
-could lead to discrimination (when
seeking insurance or medical assistance)
-fear of getting older (depression)
-creates a genetic ‘underclass’
83. Human Genome Project
A. An international cooperative venture
established to sequence the complete
human genome
B. Hope to determine the location and
structure of all genes in the human
chromosomes
C. Genome-total genetic material of a cell
D. Completed in 2000 (about 10 years
early)
84. Advantages of the Human Genome
Project
A. Understand genetic diseases
B. Figure out if any of them can be
prevented though screening
-could also be negative (in terms of
insurance or employers)
C. May lead to development of
pharmaceuticals
D. Insight into evolution and migration
patterns
85. Genetic Modification
A. Deliberate manipulation of genetic
material
B. Genetic code is universal
-it can be transferred between organisms
because the bases are the same
C. Used to create new combinations of DNA
D. Mutation and recombination occur
naturally (often are disadvantageous and
do not remain in the population)
87. Technique for gene transfer
Example-insulin production
A. Must have a vector (bacteriophage), a host
(bacteria), restriction enzymes and DNA
ligase
B. Plasmids (small circular DNA from bacteria)
are cut using restriction enzymes
C. Sticky ends are created by adding C
nucleotides
88. Technique for gene transfer
Example-insulin production
D. mRNA that codes for insulin is extracted
from the pancreas
E. Reverse transcriptase makes DNA from
the mRNA
F. Sticky ends are creating by adding G
nucleotides
G. Insulin gene and plamsid are mixed
H. They join together at the sticky ends (C
pairs with G)
89. Technique for gene transfer
Example-insulin production
I. Plasmid with the human insulin gene is
called a recombinant
J. Host cell (usually E. coli) receives the gene
K. E. coli are cultured w/ new gene
L. Insulin produced by modified E. coli is
extracted and eventually given to
diabetics
90. Other examples of genetic
modification
A. Insulin production in E. coli (discussed
above)
B. Bacteria can be modified to produce
growth hormone for cows
-cows injected with hormone increase
milk production by 10-20%
C. Breeding of plants to increase disease
resistance
D. Dog breeding
91. Benefits of genetic modification
A. Less disease (possibly)
-long term effects are unknown
B. More product
-Ex: milk
C. Some are beneficial with no know side
effects
-Ex: Insulin production
92. Negative aspects of genetic
modification
A. Introducing plasmids
-hospitals fear them
-can carry genes for antibiotic resistance
-can be passed from one species to another
B. Don’t know long term effects (Example)
-effects of using growth hormone in cows is unknown
(some could be in our food)
-cows needs extra antibiotics to stay healthy (we get
these too)
C. Could create super bacteria
D. Less than perfect becomes unacceptable (if anyone can
be genetically modified before birth)
93. Gene therapy
A. Treatment of genetic illness by
modification of genotype, or base
sequence, or allele (dominant for
recessive)
B. Best if done with stem cells
C. Many attempts have not been successful
D. Read SCID example on p. 28
94. Cloning
A. Clone-group of individuals identical in
genotype or a group of cells descended
from a single parent cell
95. How Dolly was made
1. Udder cells removed from sheep 1
a. cells grown in a culture to turn off their genes
2. Embryos removed from sheep 2
a. Nuclei removed from embryos
3. Embryos and udder cells fused by electricity to
form zygotes
4. Zygotes developed to embryos
5. Embryos implanted into Sheep 3 (the surrogate
mother)
6. Dolly develops and becomes first born clone
a. Identical to sheep 1
96. Ethical issues of cloning in humans
A. Possible to only clone a specific organ?
B. Does the whole person need to be
cloned for organ donation?
-if we take the heart, what do we do with the
rest of the body
C. Read p. 32 in the review guide for
arguments for and against therapeutic
cloning of humans.
