Name: _________________________________
Unit 7: Genetics
Part A: Fundamentals of Genetics
Gregor Mendel
o
garden peas
o
heredity
o
genetics
o
P1
o
F1
o
F2
o
dominant
o
recessive
o
hybrid
Genotype
Per. ______
Phenotype
o
homozygous
o
heterogyzous
o
alleles
o
Punnett squares
o
pollination: cross and self
Principle of Dominance and Recessiveness
Law of Segregation
Law of Independent Assortment
o
multiple alleles
o
monohybrid crosses
o
dihybrid crosses
Part B: Inheritance Pattern & Human Genetics
1. sex determination
Thomas Hunt Morgan
sex-linked genes
linkage groups
chromosome mapping
mutation
o chromosome

deletion

inversion

translocation

nondisjunction
o gene

point mutations

substitutions
 sickle-cell anemia

frame shift mutations
2. human genetics
o pedigrees
o Huntington’s Chorea
multiple alleles
polygenic traits
sex-linked
o colorblindness
o hemophilia
o Duchenne muscular dystrophy
sex-influenced traits
nondisjunction
o Down Syndrome
genetic screening/counseling
o amniocentesis
o PKU
Part C: DNA Technology
genetic engineering
restriction enzymes
plasmids
transplanting genes
recombinant DNA
transgenic organisms
DNA fingerprints
gel electrophoresis
human genome project
genetically engineered vaccines and crops
Part D: Gene Expression
gene expression
genome
repressor proteins
cell differentiation
cancer
o tumors



benign
malignant
metastasis
o carcinoma
o sarcoma
o lymphoma
o leukemia
causes of cancer
o carcinogens
o mutagens
o oncogens
o viruses
Part E: Nucleic Acid & Protein Synthesis
DNA
chromosome
nucleotide
nitrogenous base
o adenine ------ thymine
o guanine ------ cytosine
replication
transcription
RNA
o
o
o
mRNA
tRNA
rRNA
codon
protein synthesis
o translation
1
Part A: Fundamentals of Genetics
Gregor Mendel: Genetic Pioneer
genetics = study of genes
Gregor Mendel- Austrian Monk
- identified 7 pairs of contrasting characteristics in pea plants
Mendel's Experiments
developed pure strains for each of the 14 characteristics by
self-pollination (pollen from anther to stigma of same flower or plant)
then, cross-pollinated
parent plants = P1 generation
first generation offspring = F1 generation
second generation offspring = F2 generation
Mendel's Results
P1
green pods x yellow pods -----------> all green pods (F1)
F1
green pods x green pods -----------> 3/4 green pods: 1/4 yellow pods
Mendel got the same 3/4 to 1/4 ratio for each pair of pure traits that he tested
Mendel’s Conclusions
Principle of Dominance and Recessiveness
= one factor in a pair may mask the other, preventing its expression
 Incomplete Dominance- Blending of dominant & recessive traits
 Co-dominance- expression of two dominant traits
Law of Segregation & Recombination
 a pair of factors is segregated, or separated during the formation of gametes
 as a result of fertilization, alleles recombine. New allelic gene combinations are likely to be produced
Law of Independent Assortment
= factors are distributed to gametes independently of other factors
- later found to be untrue where gene linkage is present
Modern Look at Mendel's Work
Mendel's factors = genes
letters used to represent genes pairs:
-capital letter for dominant gene
-lower case letter for recessive gene
ex. tallness vs. shortness
T = tall gene
t = short gene
TT = tall
tt = short
Tt = tall
2
Genes and Appearance
genotype = gene combination present
ex. Tt, TT, tt
phenotype = physical appearance due to gene action
ex. tall, tall, short
homozygous = both genes of pair chromosomes
ex. TT, tt
heterozygous = genes of pair chromosomes
ex. Tt
- also called a hybrid
alleles = contrasting traits for same characteristic
ex. tallness vs. shortness
multiple alleles - traits with more than two alleles
ex. blood type (A, B, O)
- how many of these possible different genes can you inherit? 2
female
gametes
Punnett Squares and Probability
 show possible results of crosses
X
male
gametes
X
XX
X
Y
XX
XY
Practice:
1. cross a homozygous tall pea plant with a short pea plant.
Key: T= tall
genotype: 4Tt
t= short
phenotype: 4 tall
2. cross the hybrid F1’s from the above cross.
XY
T
TT
Tt
T
t
T
Tt
Tt
t
t
t
Tt
tt
T
Tt
Tt
genotype: 1TT: 2Tt: 1tt
phenotype: 3 tall: 1 short
3. cross a heterozygous axial-flowered pea plant with a terminal flowered pea plant.
key: A= axial
a= terminal
genotype: 2Aa: 2aa
phenotype: 2 axial: 2 terminal
A
Aa
Aa
a
a
a
aa
aa
4. cross a homozygous, round-seeded pea plant with a heterozygous round-seeded pea plant.
key: R= round
r= wrinkled
genotype: 2Rr: 2Rr
phenotype: 4 round
R
r
R
RR
Rr
R
RR
Rr
5. black coat color is dominant over white coat color in guinea pigs. Cross a
heterozygous black guinea pig with a
genotype: 2Bb: 2bb
white guinea pig.
key: B= black
phenotype: 2black: 2 white
b= white
6. cross a constricted pod pea plant with a homozygous inflated pod pea plant.
key: I= inflated
genotype: 4Ii
i= constricted
phenotype: 4 inflated
7. how would you find out if a tall pea plant is homozygous or heterozygous?
 these are all monohybrid crosses
key: T= tall
t= short
Cross with short plant to see
if you get a short offspring
t
t
B
Bb
Bb
b
b
I
Ii
Ii
i
i
T
Tt
Tt
t
tt
tt
b
bb
bb
I
Ii
Ii
3
Crosses That Involve Two Traits
= dihybrid cross
ex.
R = round
r = wrinkled
Y = yellow
y = green
 cross pure round, yellow seeded pea plants with pure wrinkled green seeded pea plants:
RRYY x rryy
 gametes = RY x ry
 F1 from above cross can produce four different gametes: RrYy x RrYy
RY
Ry
rY
ry
 Punnett square can be set up with 16 boxes
Practice
1. One reason for Mendel’s success with genetic studies of garden peas was that he
1. used only hybrid pea plants
3. studied large numbers of offspring
2. used peas with large chromosomes
4. discovered the source of variations in pea plants.
2. Each member of a pair of genes found in the same position on homologous chromosomes is known as
1. an allele
2. a gamete
3. a chromatid
4. an autosome
3. Curly hair in human beings, white fur in guinea pigs, and needlelike spines on cactus all partly describes an
organism’s
1. alleles
2. chromosomes
3. autosomes
4. phenotype
4. If one offspring is homozygous dominant and a second offspring is heterozygous for the same trait, the two
offspring would most likely
1. have the same genotype
3. have the same phenotype
2. exhibit the recessive trait
4. have different chromosome numbers
5. A student crossed wrinkled seeded (rr) pea plants with round seeded (RR) pea plants. Only round seeds were
produced in the resulting plants. This illustrates the principles of
1. independent assortment
2. dominance
3. segregation
4. incomplete dominance
6. A trait that is not visible in either parent appears in several offspring. Which genetic concept does this
demonstrate?
1. linked genes
2. segregation
3. replication
4. sex determination
7. Sexually reproducing species show great variation than asexual reproducing species because of
1. lower rates of mutation
3. the occurrence of polyploidy
2. environmental changes
4. the recombination of alleles
8. Which statement describes how two organisms may show the same trait, yet have different genotypes for that
phenotype?
1. One is homozygous dominant and the other is heterozygous.
2. Both are heterozygous for that trait.
3. One is homozygous dominant and the other homozygous recessive.
4. Both are homozygous for that dominant trait.
9. Two pea plants hybrid for a single trait produced 60 pea plants. Approximately how many of these pea plants are
expected to exhibit the recessive trait?
1. 15
2. 30
3. 45
4. 60
10. In guinea pigs, the gene for black coat color is dominant over the gene for white coat color. In a cross between
two hybrid black guinea pigs, what percentage of the offspring is likely to have the same coat color as the parents?
1. 25%
2. 50%
3. 75%
4. 100%
4
Genetics Crosses Practice
1. Write the genotypes for pea plants with the following phenotypes:
a. homozygous tall
TT
b. short
tt
c. heterozygous tall
Tt
2. Write the phenotypes of the pea plants with the following genotypes:
a. Tt tall
b. TT tall
c. tt
short
3. Write the genotypes for pea plants with the following phenotypes:
a. heterozygous axial Aa
b. homozygous axial
AA
c. terminal flowers
aa
4. Write the phenotypes for the following genotypes:
a. Aa axial
b. AA axial
c. aa
terminal
Keeping in mind the following crosses are examples of incomplete dominance, complete the following crosses:
5. Cross a red four o’clock flower with a pink four o’clock flower.
R
R
r
R
RR
RR
Rr
Rr
F1
genotypes: 2RR: 2Rr
phenotypes: 2red: 2 pink
6. Cross a white andalusian chicken with a black andalusian chicken.
W
w
w
W
Ww
Ww
Ww
Ww
F1
genotypes: 4 Ww
phenotypes: 4 blue
7. Cross two roan short-horned cattle.
R
R
r
r
RR
Rr
Rr
rr
F1
genotypes: 1RR: 2Rr: 1rr
phenotypes: 1 red: 2 roan: 1 white
5
Punnett Square Worksheet
Complete the following monohybrid crosses: draw a Punnett square, list the ratio and describe the offspring. Be
sure to remember that the capital letter is dominant.
Example:
A green pea plant (GG) is being crossed with a green pea plant (Gg) yellow is the recessive color.
G
G
g
GG
Gg
G
Genotype=
GG
Gg
Phenotype= 4 Green pea plants: 0 yellow pea plants
G
1. A green pea plant (Gg) is crossed with a yellow pea plant (gg).
genotype= 2Gg: 2gg
phenotype= 2 green: 2 yellow
2. A tall plant (TT) is crossed with a tall plant (Tt).
genotype= 2TT: 2Tt
phenotype= 4 tall
2 GG: 2 Gg : 0 gg
g
g
Gg
gg
g
Gg
gg
T
T
T
TT
TT
t
Tt
Tt
T
t
Tt
tt
Tt
tt
3. A tall plant (Tt) is crossed with a short plant (tt).
t
genotype= 2Tt: 2tt
phenotype= 2 tall: 2 short
t
R
4. A red flower (Rr) is crossed with a white flower (rr).
genotype= 2 Rr: 2 rr
phenotype= 2 red: 2 white
r
Rr
rr
r
Rr
rr
r
r
r
rr
rr
r
rr
rr
5. A white flower (rr) is crossed with a white flower (rr).
genotype= 4 rr
phenotype= 4 white
6. A black chicken (BB) is crossed with a black chicken (BB).
genotype= 4 BB
phenotype= 4 black
6
r
B
B
B
BB
BB
B
BB
BB
Punnett square problems continued
Complete the following problems. List the parent genotypes, draw and fill in a Punnett square, and then list the
offspring genotypes and phenotypes.
1.
B
b
A homozygous dominant brown mouse is crossed with a heterozygous brown mouse (tan is the recessive
color).
B
B
BB
BB
Bb
genotype= 2 BB: 2 Bb
phenotype= 4 brown
Bb
2. Two heterozygous white (brown fur is recessive) rabbits are crossed.
W
w
WW
W
Ww
w
Ww
genotype= 1WW: 2Ww: 1ww
phenotype= 3 white: 1 brown
ww
3. Two heterozygous red flowers (white flowers are recessive) are crossed.
R
R
r
r
RR
Rr
genotype= 1RR: 2Rr: 1rr
phenotype= 3 red: 1 white
Rr
rr
4. A homozygous tall plant is crossed with a heterozygous tall plant (short is the recessive size).
T
T
t
T
TT
TT
Tt
Tt
genotype= 2 TT: 2Tt
phenotype= 4 tall
5. A heterozygous white rabbit is crossed with a homozygous black rabbit.
W
w
w
w
Ww
ww
Ww
ww
genotype= 2 Ww: 2 ww
phenotype= 2 white: 2 black
7
Part B: Inheritance Patterns and Human Genetics
1. Chromosomes and Inheritance


sex determination
o Thomas Hunt Morgan
o used Drosophila melanogaster
= fruit flies
o why did he use fruit flies?
 only 4 pairs of chromosomes
 they are very large & easy to see
 they breed fast
o of 4 pairs of chromosomes, one pair different in males than in females
 females: two chromosomes identical
 males: one chromosome looked like female and the other was shorter and hook shaped
o Morgan called these sex chromosomes
XX (female)
XY (male)
Cross a male with a female and give genotype and phenotype ratios:
X
X
X
XX
XX
Y
XY
XY
Are Chromosomes the Same in Male and Female?
Chromosomes in Female and Male.
X
X
X
Y
1. How many chromosomes are in the body cells of a male fruit fly? (Note: the large dots are chromosomes, too.)
______________________________________________8_______________________________________
2. How many chromosomes are in the body cells of a female fruit fly? 8
3. In the male fruit fly, how many of the pairs consist of two chromosomes that look alike? 3
4. In the female fruit fly, how many of the pairs consist of two chromosomes that look alike? 4
5. What names are given to each of the two unlike chromosomes in the male? XY
6. What is the name given to the two similar corresponding chromosomes in female? XX
8
How Sex Is Decided
Complete the following statements.
1. The number of chromosomes in the body cells of a fruit
fly is 8.
2. The male fruit fly has two sex chromosomes, called
X and Y.
3. The female fruit fly has two sex chromosomes, called X
and X.
4. A male fruit fly makes two kinds of sperm cells; one kind
has the X chromosome, the other has theY.
5. The sex chromosome found in every egg is the
X chromosome.
6. When an egg is fertilized, a female will result if the
chromosome combination is XX.
7. A male will result if the chromosome combination is
XY.
8. In any population, the proportion of males to females is
about 50/50 .
9. The sex of the offspring is decided at the moment when
the egg is fertilized.
10. The sex in humans and fruit flies that has unlike sex
chromosomes is the male.


Chromosome map: diagram showing the location of genes on a chromosome
o the farther apart two genes are, the more likely they will be separated by crossing over
Mutations: a change in DNA
o Germ cell mutation: occurs in the organism’s gametes (sex cells); do not affect the organism; may be
passed on to offspring
o Somatic mutations: body cell mutations; will affect the organism; not passed on; two types:
 Chromosome Mutations: change structure of a chromosome or loss of entire chromosome
 deletion: loss of a piece of chromosome
 inversion: segment breaks off and reattaches in reverse
order
 translocation: segment breaks off and attaches to
another non-homologous chromosome
 non-disjunction: failure of a chromosomes to
separate correctly
ex. extra chromosome or lacks a chromosome
 Gene Mutations: change in DNA of a single gene
 point mutation: substitution, addition or deletion of
1 nucleotide of a codon
 substitution: one nucleotide replaced by a different
nucleotide
ex. sickle cell- T is substituted by A; causes defective
hemoglobin; sickle shaped red blood cells
 frame shift mutation: occurs when an addition or
deletion causes the shifting of the group of three
making a codon
9
2. Human Genetics
difficult to study. Why? We grow over very long period of time; harder to take care of; small amount of
offspring.
 Pedigree analysis
o pedigree: family record that shows how a trait is inherited over generations
 Genetic traits and disorders
o Single-allele traits:
 Dominant:
 Huntington’s Disease: forgetfulness and irritability in 30’s – 40’s; loss of muscle
control, spasms, mental illness, death
 Achondroplasia: dwarfism
 Polydactyly: many fingers and toes
 Cataracts: clouding of the lens of the eye
 Recessive:
 Albinism: lack of pigmentation in hair, skin, eyes
 Cystic Fibrosis: abnormal cellular secretions of thick mucus which accumulates in
lungs
 Phenylketonuria (PKU): inability to metabolize phenylalanine in milk; PP and Pp = normal;
pp= PKU
 build-up causes mental retardation
 babies tested; those with PKU not given phenylalanine in diet
 Tay-Sachs Disease: causes death by deteriorationfrom lack of enzyme to breakdown
fatty deposits on nerve and brain cells
o Multiple Alleles: traits controlled by 3 or more alleles of the same gene
 ABO blood groups controlled by 3 alleles: IA; IB; ii
 each person’s blood contains 2 of these alleles
 IA, IB are codominant (both expressed when together) and are both dominant to the i
 A type = IA IA; IAi
 B type = IB; IB i
 AB type = IA IB
 O type = ii
 Which cross could result in all four blood types in the offspring? Show results of cross below:
IA
i
IA IB
IB
i
10
IB i
IAi
ii
Pedigree Studies
Pedigrees are not reserved for show dogs and race horses. All living things, including humans, have
pedigrees. A pedigree is a diagram that shows the occurrence and appearance, or phenotype, of a
particular genetic trait from one generation to the next in a family. Genotypes for individuals in a
pedigree usually can be determined with an understanding of inheritance and probability.
In this investigation, you will
(a) Learn the meaning of all symbols and lines that are used in a pedigree.
(b) Calculate expected genotypes for all individuals shown in pedigrees.
Procedure
Part A. Background information
The pedigree in Figure 20-1 shows the pattern of the
inheritance in a family for a specific trait. The trait being
shown is earlobe shape. Geneticists recognize two general
earlobe shapes, free lobes and attached lobes (Figure 202). The gene responsible for free lobes (E) is dominant over
the gene for attached earlobes (e).
In a pedigree, each generation is represented by a
roman numeral. Each person in a generation is numbered.
Thus, each person can be identified by a generation numeral
and individual number. Males are represented by squares
whereas females are represented by circles.
Part B. Reading a Pedigree
In Figure 20-1, person’s I-1 and I-2 are the parents. The
line which connects them is called a marriage line. Person’s
II-1, 2 and 3 are their children. The line which extends
down from the marriage line is the children line. The
children are placed left to right in order of their births.
That is, the oldest child is always on the left.
1. What sex is the oldest child? female
2. What sex is the youngest child? male
Using a different pedigree of the same family at a later time
shows three generations. Figure 20-3 shows a son-in-law as
well as a grandchild. Generation I may now be called
grandparents.
3. Which person is the son-in-law? II-1
4. To who is he married? II-2
5. What sex is their child? female
11
Part C. Determining Genotypes from a Pedigree
The value of a pedigree is that it can help
predict the genes (genotype) of each person for
a certain trait.
All shaded symbols on a pedigree
represent individuals who are homozygous
recessive for the trait being studied.
Therefore, person’s I-1 and II-2 have ee
genotypes. They are the only two individuals who
are homozygous recessive and show the
recessive trait. They have attached earlobes.
All un-shaded symbols represent
individuals who have at least one dominant gene.
These persons show the dominant trait.
To predict the genotypes for each person
in a pedigree, there are two rules you must
follow.
To determine the second gene for the
persons who show the dominant trait, a Punnett
square is used. In Figure 20-4, we already know
that the grandfather (I-1) is ee, if the
grandmother (I-2) were EE; could any ee
children like (II-2) be produced? A Punnett
square shows this combination to be impossible.
Thus, the grandmother must be heterozygous of
Ee.
6. (a) Do the following Punnett squares to show
the possible outcomes of persons I-1 & I-2.
(b) Can an Ee parent and an ee parent have
Rule 1: Assign two recessive genes to any person
on a pedigree whose symbol is shaded. (These
persons show the recessive trait being studied.)
Small letters are written below the person’s
symbol.
Rule 2: Assign one dominant gene to any person
on a pedigree whose symbol is un-shaded. (These
persons can show the dominant trait being
studied.) A capital letter is written below the
person’s symbol.
These two rules allow one to predict some of the
genes for the persons in a pedigree. Figure 20-4
shows the genes predicted by using these two
rules.
the results in generation II? yes
(c) Can an EE parent and an ee parent have
the results shown in Generation II? no
7. (a) Predict the second gene for person II-3.
(Read the Punnett square.) e
(b) Predict the second gene for persons II-4.
e
(c) Could child II-3 or II-4 be EE? no
Explain. e from father
12
To predict the second gene for person II-1, a
different method must be used, since he could
be either EE or Ee.
8. (a) Do the following Punned squares to show
the possible outcomes of persons II-1 &
II-2.
Examine the pedigree:
(b) Can and EE person married to an ee
person (II-2) have children with free earlobes?
yes- all
(c) Can an Ee person be married to an ee
person have children with free earlobes?
yes- 50%
9. (a) Which Punnett square, A, B, or C, would
best fit this family? B
(b) Explain. offspring with ee
In this case, the second gene from person
II-1 cannot be predicted using Punnett squares.
Either genotype Ee or EE may be correct. When
this situation occurs, both genotypes are written
under the symbol (Figure 20-5)
Predicting the second gene for III-1 results
in her being heterozygous. Although her mother
must provide her with one recessive gene, she
has free lobes, so the second gene must be
dominant (Figure 20-5).
At some time in the future, if II-1 and
II-2 have many more children, one might be able
to predict the father’s second gene. For
example, if they have ten children and all show
the dominant free lobes, one could safely
conclude that he is EE. If, however, they have
some children with attached earlobes (ee), then
he must be Ee
When both parents show a dominant trait
and their child or children all show a dominant
trait, one cannot predict the second gene for
anyone if only a small family is available.
Analysis
1. Draw a pedigree for a family showing two
parents and four children.
(a) Include a marriage line and label it.
(b) Include a children’s line and label it.
(c) Make the oldest two children boys and the
youngest two girls.
2. Fill out the following pedigree. Find the
genotype for each person. Use B & b.
13
Knowing Your Blood Type
If you ever have surgery and need blood, your doctor will need to know your blood type. The only type of blood you
can receive is blood that will not clot with your blood.
1. Fill in the blanks in the table.
If you have blood type
your genes are
A
IA IA
or
IAi
IB IB
or
B
IBi
so you received this gene
from one parent
and this gene from the
other parent
IA
IA
IA
i
IB
IB
IB
i
AB
IA IB
IA
IB
O
ii
i
i
2. Use the information below to ill in the table.
 Type A has plasma proteins that clot with red cell proteins from donor type B.
 Type B has plasma proteins that clot with red cell proteins from donor type A.
 Type AB has no plasma proteins that will clot with red cell proteins from any donor types.
 Type O has plasma proteins that clot with red cell proteins from donor types A, B, or AB.
Blood Type
Can receive blood from type(s)…
A
A, O
B
B, O
AB
A, B, AB, O
O
O
Universal Donor = O
Universal Recipient= AB
14
Sample Questions
1) Brad is Type A blood and Angelina is Type B blood. Assume they are both homozygous.
What are the possible phenotypes of their children?
IA
IA
IB
IA IB
IA IB
IB
IA IB
IA IB
2) Jennifer is Type O. If Brad and Jennifer had a child when they were married what could have been the
phenotypes of their children?
i
i
IA
IA i
IA i
IA
IA i
IA i
3) Could a man with type B blood and a woman with type AB produce a child with type O blood?
IB
IA
IB
IB
IA IB
IA IB
IB IB
IB IB
IB
IA
IB
i
IA IB
IA i
IB IB
IB i
15

Sex Determination
* It is the father's sperm that determines the sex of the baby
 Sex Linkage:
o more genes are carried on the sex chromosome = X-linked genes
o
sex-linked genes are on one chromosome; these are linked
= get inherited together


o
Colorblindness: recessive; inability to distinguish colors (red/green)
Hemophilia: recessive; bleeder’s disease; impaired ability of blood to clot
 Duchenne Muscular Dystrophy: weakens and destroys muscle tissues
representing sex-linked traits:
 XC = normal, Xc = colorblind

XH = normal, Xh = hemophiliac
 X = normal,
Colorblindness Example:
X = disease trait
normal female
normal female
colorblind female
normal male
colorblind male
o
16
carrier: heterozygous female
 she has one copy of the trait but does not have the disease
 she can pass the trait to her children
 men can NEVER be carriers - if they have one copy of the allele they automatically have the
disease
Figure 1: Inheritance of Hemophilia
“Carrier” Mother and Father Without Hemophilia
Figure 2: Inheritance of Hemophilia
Mother who is not a carrier and Father With Hemophilia
Answer the following according to figure 1.
1. The mother has a gene for bleeding but she is not a “bleeder” because this gene is recessive.
2. The gene for bleeding in the X chromosome passes from the mother to one of the daughters and one of the
sons.
3. The daughter who receives the X chromosome is, like her mother, a carrier.
4. The son who receives the X chromosome is, on the other hand, a hemophiliac.
5. This happens because in the male, the Y chromosome has no gene for clotting to dominate the gene for bleeding.
6. In other words, any male having a single gene for bleeding will be born with hemophilia.
Sample problems:
1) Wilma, a colorblind woman, marries Fred, a normal male.
a) What percent of their children will be colorblind?
50%
X
X
XX
X
XX
b) What percent of their sons?
100%
c) What percent of their daughters?
0%
Y
XY
X
2) Neither Marie or Frank is colorblind but their son Raymond is
colorblind.
How is this possible?
Marie is a carrier
XY
X
XX
XX
XY
XY
X
Y
17

Sex-Influenced Traits: male or female hormones may influence gene expression
ex. baldness controlled by gene B; dominant in males but recessive in females
BB= bald in male and female
Bb= bald in male, normal in female; caused by testosterone

Polygentic Traits:
ex. skin color is influenced by 3-6 genes; control the amount of pigment (melanin) in the skin

Non-Disjunction Disorders: failure of chromosomes to separate in meiosis
o
Monosomy X aka Turner’s syndrome: 45 chromosomes; underdeveloped, sterile female
o
Trisomy X: XXX; super females; some retarded
o
Klinefelter’s syndrome: XXY; normal egg x XY sperm
 sterile, underdeveloped males
XYY: tall aggressive males, criminals?
Down Syndrome: Trisomy 21
 extra chromosome 21
 mild to severe retardation, facial features, muscle weakness, heart defects, short stature
o
o
Monosomy X

Trisomy X
Detecting Human Genetic Disorders
o Genetic screening: exam of person’s chromosomes
ex. karyotype: picture of chromosomes
blood test
amniocentesis: testing of amniotic fluid from embryo
chorionic villi sampling: sample tissue between
mother’s uterus and placenta
o
Genetic counseling: talk to patients about genetic
disorders and risks of having affected children
Human Characteristics
18
Trisomy 21
As you know, chromosomes work in pairs. The members of each chromosomes pair are called homologous
chromosomes, and the two chromosomes of each pair are approximately the same length, the same shape, and carry
alleles for the same genes. Each chromosome of a pair comes from a different parent: one from the mother
through the egg and the other from the father through the sperm. Humans have 23 pairs of chromosomes, or a
total of 46 chromosomes, per cell. Only 22 of these pairs are truly homologous. The twenty-third pair, the sex
chromosomes, may or may not match, depending on whether the individual is female (XX) or male (XY).
The diagram below shows a generalized view of one pair of homologous chromosomes from an individual
human. Assume the chromosome on the left is from the father and the one on the right is from the mother.
You should be aware that any two alleles of a gene have different effects on the trait that they control.
For example, the M allele causes more melanin to be made in the skin, giving it a darker color; the m allele causes
less melanin, resulting in a lighter skin color. Even though the effects are different, the two alleles in an allele pair
control the same trait- in this case, skin color.
Sample Problem
Determine the genotypes and corresponding phenotypes of the person whose chromosomes are shown above.
Genotypes
Phenotypes
Ff
freckled
Tt
taste ability
Bb
brown hair
Hh
has disease
Rh+ RhRh+ (positive is dominant)
Mm
darker skin
Cc
normal vision
Which parent donated a blond hair allele?
mother
Was that parent blond?
possibly- no way to know for sure from information given
Is this person blond?
no
Is this person colorblind?
no
Can this person taste the chemical?
yes
Is his person freckled?
yes
From the information provided, can you determine
no- these are not the sex chromosomes and the genes shown
this person’s eye color or sex?
do no control eye color
Exercises
19
Determine the genotypes and corresponding phenotypes of the people whose chromosomes appear below. Then
answer the questions that follow.
Genotypes
Phenotypes
1.
a.
Ff
freckled
b.
tt
can’t taste
c.
Bb
brown hair
d.
hh
no disease
e.
Rh+Rh+
f.
mm
light skin
g.
CC
normal vision
Genotypes
Phenotypes
Rh+
2.
a.
Ff
freckled
b.
TT
can taste
c.
bb
blond hair
d.
hh
no disease
e.
Rh+Rh-
f.
mm
light skin
g.
Cc
normal vision
Genotypes
Rh+
Phenotypes
3.
20
a.
ff
unfreckled
b.
tt
can’t taste
c.
BB
brown hair
d.
HH
has disease
e.
Rh-Rh-
f.
MM
dark skin
g.
cc
colorblind
Rh-
4. Was Kara’s father freckled?
4. can’t tell
5. Was Kara’s mother freckled?
5. yes
6. Which of the two genotypes, Ff or FF,
would have more freckles?
6. the same
7. Did either of Kara’s parents have taste ability?
7. can’t tell
8. Are all of Juanita’s brothers and sisters blond?
8. can’t tell
9. When Juanita has children of her own, will they
have taste ability?
9. yes (TT)
10. What color is Phil’s hair?
10. brown
11. Is Phil skin color light or dark?
11. dark
12. Were both Phil’s parents colorblind?
12. can’t tell
Part C: DNA Technology
1. The New Genetics


DNA technology can be used to cure diseases, make better crops, animals, or drugs
Manipulating Genes:
o to isolate and transfer specific DNA segments, restriction
enzymes are used to cut a piece of
DNA
o single chains of DNA are created with “sticky - ends”
o
sticky ends bind to complementary sticky ends form
recombinant form out of DNA from 2
organisms
cloning vectors: carrier used to clone a gene and transfer it to another organism
o
plasmid: ring of DNA in a bacterium
o

Transplanting Genes:
o plasmids are used to clone a gene so that bacteria will produce a specific protein
ex. insulin
o
Steps:
1. restriction enzymes cut the segment of DNA from a human cell that contains the insulin gene and
the circular plasmid in the bacteria
2. recombinant DNA is formed by combining the human and bacterial DNA segments
3. The loop of DNA is inserted into a bacterial cell
4. The bacterial cell will produce the insulin and be duplicated every time it divides
o transgenic organism: host receiving the recombinant DNA
21
2. DNA Technology Techniques

DNA Fingerprints:
o patterns of bands that make up fragments from an individual’s DNA
Uses:
 comparing different species to determine how closely related


compare blood, tissue at blood scene with a suspect’s blood
 establish relatedness or paternity
Making a DNA Finger Print
1. DNA segment cut into pieces by restriction enzymes
2. gel electrophoresis: separates the DNA fragments by size
and charge
 DNA fragments are placed in wells in a gel

an electric current is run through
the gel

DNA fragments (- charge) migrate to the positive
charged end of gel, not at the same rate

 smaller fragments migrate faster
3. Make visible only bands being compared by using radioactive probes and
photographic film
Human Genome Project:
o Goals:
 determine the nucleotide sequence of the entire human genome


3 billion nucleotide pairs, 100,000 genes
map the location of every gene on each chromosome
3. Practical Uses of DNA Technology



Producing Pharmaceutical Products: that are safer and less expensive than produced by conventional means
Genetically Engineered Vaccines:
o vaccine: solution containing a harmless version of virus or bacterium
o new DNA technology can prevent a pathogen (disease causing agent) from harming someone that has
received a vaccine (only rare cases)
Increasing Agricultural Yields:
o produce bacteria- resistant, herbicide- resistant, and insect - resistant crops
o improves the quality and quantity of the human food
o isolate genes from nitrogen-fixing bacteria and transplant to plants so they can be grown in nitrogenpoor soil without fertilizers
Part D: Gene Expression
1. Control of Gene Expression

activation of a gene that results in the formation of a protein


when transcription occurs a gene is “expressed” or “turned-on”
ex. gene for blue eyes is “expressed” only in the iris of the eye
genome: the complete genetic material contained in an individual

repressor protein: inhibits a gene from being expressed (“turns off the gene”)
2. Gene Expression and Development

Cell Differentiation: development of cells with specialized-functions
o
22
controlled by gene expression


Cancer:
o tumor: abnormal group of cells from uncontrolled, abnormal cell division
benign: no threat unless compressing a vital organ; in a single mass
ex. fibroid cyst in uterus or breast, warts

malignant: abnormal cells invade and destroy healthy tissue elsewhere in body
 metastasis: spreading of cancer beyond original site
Kinds of Cancer:
o carcinoma: skin, lining of organs
ex. lung cancer, breast cancer
o
o


sarcomas: bone and muscle tissue
lymphomas: solid tumors in blood-forming tissue and may cause leukemia
o leukemia: uncontrolled production of white blood cells
Causes of Cancer:
o mutations that alter expression of genes
 spontaneous
 caused by carcinogen (substance that increases the risk of cancer)
ex. smoking, asbestos, radiation
 viruses (HPV)
o mutagen: agents that cause mutations
Plant and Animal Breeding
Since the beginning of agriculture, farmers have tried to improve plants and animals for food. They have
mated animals with the traits they wanted. They have saved seeds from the best plants and planted the seeds the
following year. Selecting hardy, productive plants and animals to produce the next generation is called selective
breeding. Selective breeding is based on the techniques of mass selection, inbreeding, and hybridization.
Mass selection is choosing the best plants and animals from a large number for further breeding.
Inbreeding is the mating of related individuals to establish pure lines. Breeders use inbreeding to keep desirable
traits in a plant or animal species. In hybridization, two different but related varieties or species of plants or
animals are mated. The results of the mating are called hybrids.
Examples of selective breeding are described below. In the space provided, identify each example as
either mass selection, inbreeding, or hybridization.
1. Luther Burbank grew large numbers of fruits, flowers, vegetables and grains. He selected the best plants to
breed for the next generation. He grew thousands of plants trying to produce one improved species.
Mass selection
2. A dog had five puppies. Two of the puppies, a male and a female, were a golden color. When the puppies were
mature, they were mated to each other in an attempt to produce puppies with the same golden color.
inbreeding
3. A male championship race horse was mated to its female offspring to produce a horse with the championship
traits of the male horse. inbreeding
4. An apple liked by consumers for its taste, crispness, and long storage life was pollinated with the pollen of an
apple that resisted disease. The resulting apple has the taste, crispness, and long storage life shoppers want. It
also resists diseases. hybridization
5. A large pink daisy is discovered in a bed of all white daisies. Seeds from the pink daisy are collected and planted
the next year with hopes of producing more pink daisies. Mass selection
6. The bull mastiff is the result of breeding an aggressive bull dog with a large, fast, gentle mastiff. The bull
mastiff has proved to be a good guard dog. hybridization
23
Practice Questions
1. Which may occur in meiotic division 1 of a primary sex cell?
1. fertilization
2. crossing over
3. polyploidy
4. differentiation
2. Cosmic rays, x-rays, ultraviolet rays, and radiation from radioactive substances may function as
1. pollinating agents
2. plant auxins
3. mutagenic agents
4. animal pigments
3. In a particular variety of corn, the kernels turn red when exposed to sunlight. In the absence of sunlight, the
kernels main remain yellow. Based on this information, it can be concluded that the color of these corn kernels is
due to the
1. effect of sunlight on transpiration
3. principle of sex linkage
2. law of incomplete dominance
4. effect of environment on gene expression
4. The Himalayan rabbit has white fur over most of its body, but it has black fur on its tail, ears, and tips of its
legs and nose. Two rabbits that are homozygous for this hair pattern are mated. When their offspring are exposed
to normal temperatures, they exhibit normal Himalayan hair pattern. However, when their offspring are exposed to
low temperatures (10oC), they have black fur covering their entire bodies. This illustrates:
1. that mutations are caused by heat radiation
2. that the traits of an organism are determined by its genes
3. the importance of the environment in gene expression
4. the law of incomplete dominance
5. Substances that cause a chemical change in the DNA of a cell are known as
1. glycogens
2. chromatids
3. mutagens
4. chromosomes
6. Race horses show many variations from the wild horses ancestors from which they were derived. It is most likely
that these variations between race horses and their ancestors are due to
1. use and disuse or organs
3. gene cloning
2. artificial selection
4. chromosomal non-disjunction
7. The process by which homologous chromosomes exchange segments of DNA is
1. segregation
2. fertilization
3. crossing over
4. independent assortment
8. During egg cell production in a human female, the 21st pair of chromosomes may fail to separate. This failure to
separate is known as
1. crossing over
2. polyploidy
3. gene mutation
4. non-disjunction
9. A common practice used by breeders to maintain a desired trait in dogs is
1. artificial selection
2. regeneration
3. vegetative propagation
4. sporulation
10. Strontium-90, a radioactive isotope found in nuclear fallout, is incorporated and used by the human body in
much the same manner as calcium. Because of its radioactive nature, strontium-90 would probably
1. cause disjunction of chromosomes
3. act as a mutagenic agent within bone cells
2. take the place of phosphorus in chromosomes
4. inhibit the development of mutations
11. Although genetic mutations may occur spontaneously in organisms, the incidence of such mutations may be
increased by
1. radioactive substances in the environment
3. lack of vitamins in the diet
2. a long exposure to humid climates
4. a short exposure to freezing temperatures
24
Part F
Nucleic Acid and Protein
Synthesis
DNA
proteins are found in all living things
are species specific/individual specific
(transplant rejections)
Importance of DNA
chromosomes: DNA + RNA
nucleotide= 5 carbon sugar, nitrogen base and
phosphate
draw nucleotide here:
1953 Watson and Crick- model of DNA molecule
as a spiral helix
Structure of DNA
= double helix (like ladder twisted on its long
axis)
side of ladder - sugar and phosphate
rungs = nitrogen base pairs
only _________ nitrogen bases in DNA:
purines:
adenine and guanine
pyrimidines:
thymine and cytosine
arranged in complementary base pairs:
A----T
all
G----C
go
The Roles of DNA and RNA
teachers
crazy
Replication of DNA
replication= The process of making an identical copy of a section of
double stranded DNA
DNA ladder unzips at base pairs
free nitrogen bases assemble on the open strands, forming two new complete strands
occurs with great degree of accuracy
requires enzymes called DNA polymerases
25
RNA
Transcription: Copies the Information of DNA
transcription = making complementary copy of DNA segment (is it an exact copy? Explain why or why not)
made inside nucleus
leaves nucleus and goes to ribosome in cytoplasm to direct synthesis of protein
RNA differs from DNA
RNA: -one strand
-ribose sugar
-uracil replaces thymine as fourth nitrogen base
-is smaller than DNA
three types of RNA:
1/ mRNA - carries "recipe" for protein to ribosome
2/ tRNA - carries amino acids to ribosomes
3/ rRNA - in ribosomes
mRNA transcribed in nucleus:
mRNA leaves nucleus and moves to ribosome to direct protein synthesis
Codons Determine Amino Acid Sequence
codon = group of three nitrogen bases on mRNA that codes for an amino acid
-also called triplets
2 to 3 codons for each a.a. (20-22)
1 codon (AUG) is start codon
several stop codons
Protein Synthesis
Translation: Proteins are Synthesized
mRNA moves to ribosome
acts as template for protein synthesis
tRNA has anticodon that will match mRNA codons
as ribosomes move across mRNA, tRNA brings appropriate amino acids to it
amino acids bond together forming a peptide chain, which forms a polypeptide, which, when reaching
macromolecular size, is a protein (polypeptide)
26
Practice Questions:
1. Which is a five carbon sugar found in an RNA molecule?
1. uracil
2. ribose
3. adenine
4. glucose
2. Which base is found in DNA but not in RNA?
1. adenine
2. cytosine
4. uracil
3. thymine
3. The sequence of nucleotides in a messenger RNA molecule is determined by the sequence of nucleotides in a
1. transfer RNA molecule
3. polysaccharide molecule
2. protein molecule
4. DNA molecule
4. In protein synthesis, the code for a particular amino acid is determined by
1. the one gene-many enzyme hypothesis
3. multiple alleles
2. a sequence of three nucleotides
4. the number of messenger RNA molecules
5. If the code for glutamic acid is ATG on the DNA molecule, this code on the transfer RNA molecule may be
written as
1. ATG
2. CTG
3. AUG
4. GTA
6. To which organelles is messenger RNA attached?
1. chloroplast
2. ribosomes
3. mitochondria
4. vacuoles
7. During protein synthesis, amino acids are picked up in the cytoplasm and positioned at the ribosomes by
1. unattached nucleotide molecules
3. molecules of DNA
2. polypeptide molecules
4. molecules of RNA
8. In protein synthesis, which sequence of bases in transfer RNA will pair up with the sequence GGU found in the
messenger RNA?
1. CCT
2. CCA
3. AAC
4. UUA
9. Which is the correct sequence of code transfer involved in the formation of a polypeptide?
1. DNA, tRNA, mRNA
3. mRNA, tRNA, DNA
2. tRNA, DNA, mRNA
4. DNA, mRNA, tRNA
10. The position of an amino acid in protein molecule is determined by the
1. concentration of amino acids in the cytoplasm
2. amount of ATP in the cell synthesizing the protein
3. sequence of nitrogenous bases in DNA
4. sequence of amino groups in the amino acid
11. A sequence of three nitrogenous bases in an mRNA molecule is known as
1. codon
2. gene
3. polypeptide
4. nucleotide
12. In the synthesis of proteins, what is the function of mRNA molecules?
1. They act as a template for the synthesis of DNA.
2. They carry information that determines the sequence of amino acids
3. They remove amino acids from the nucleus.
4. They carry specific enzymes for dehydration synthesis.
13. A change in the base sequence of DNA is known as
1. a gene mutation
2. a karyotype
3. nondisjunction
4. polyploidy
14. Which nuclear process is represented below?
1. recombination
2. fertilization
3. replication
4. mutation
27
Protein Synthesis
Protein synthesis is a complex process. You will trace the steps that are involved in the protein synthesis of a part
of a molecule of oxytocin. Oxytocin is the pituitary hormone that helps regulate blood pressure, stimulates the
uterus to contract during childbirth, and stimulates the production of milk after childbirth.
A. Protein synthesis begins with DNA in the nucleus. Below is a DNA sequence that could code for part of a
molecule of oxytocin.
 Write the sequence of messenger RNA (mRNA) codons that would result from the transcription of this
portion of DNA . The arrow marks the starting point
(Nucleus) DNA
TRANSCRIPTION:
mRNA: (Codon)
ACA - ATA - TAG - CTT - TTG - ACG - GGG - AAC - CCC - ATT
1
2
3
4
5
6
7
8
9
10
1
UGU
2
3
4
5
– UAU - AUC - GAA – AAC -
6
7
UGC – CCC -
8
UUG
9
10
- GGG - UAA
B. After transcription (in nucleus), mRNA attaches to a ribosome where translation (in cytoplasm) takes place. Each
codon of mRNA bonds with an anticodon of a transfer RNA (tRNA) and each tRNA molecule bonds with a specific
amino acid. The table below shows the mRNA codons and the amino acids for which they code. For example, if you
were given the codon AGA, you can see from the table that these bases code for the amino acid arginine.
Use the information on the codon chart to fill in the following table. The first row has been completed to get you
started.
DNA triplet
AAA
GTC
ACT
UAC
mRNA
1. UGU
2. UAU
3. AUC
4. GAA
5. AAC
6. UGC
7. CCC
8 UUG
9. GGG
10. UAA
28
mRNA codon
UUU
CAG
UGA
AUG
tRNA codon
AAA
GUC
ACU
UAC
Amino Acid
Phenylalanine
glutamine
STOP
Methionine (START)
Use the mRNA sequence from A to write the sequence (1 –10) of amino acids in this part of the oxytocin molecule.
Cysteine - Tyrosine - Isoleucine – Glutamic Acid – Asparagine – Cystein – praline – leucine – glycine - STOP
1. How many amino acids make up this portion of the oxytocin molecule? 9 + 1 stop codon
2. What is the purpose of the UAA codon? tells the ribosome to stop protein synthesis
C. In order to get another view of the entire process of protein synthesis, label the structures on the diagram
below. (ribosome (rRNA), mRNA [codon=3 bases], tRNA [anticodon=3 bases], protein, DNA
DNA
mRNA
tRNA
mRNA
ribosome
To complete the chart below, give the name and a brief description of each step in protein synthesis that occurs in
the part of the cell shown in the diagram above.
PART OF CELL
Name of Protein Synthesis Process
Description
Nucleus
transcription
ribosome
translation
RNA code  amino acids
cytoplasm
protein modified
polypeptide chains folded to into proteins
DNA  mRNA
29
Base your answers to questions 1 through 3 on the diagram below showing a joining of two amino acids that occurs
within cells.
1. The process represented in the diagram occurs on the cell organelle known as a
1. vacuole
2. ribosome
3. chloroplast
4. mitochondrion
2. Which amino acid would be transferred to the position of codon CAC?
1. leucine
2. glycine
3. valine
4. histidine
3. The process represented in the diagram is
1. lipid digestion
2. cell respiration
3. protein synthesis
4. protein hydrolysis
4. Proteins are made from amino acids by the process of
1. hydrolysis
2. pinocytosis
3. active transport
4. dehydration synthesis
Use the following choices for questions 5 through 8.
Types of Nucleic Acid Molecules
(1) DNA molecule, only
(2) RNA molecules, only
(3) Both DNA and RNA molecules
(4) Neither DNA nor RNA molecules
5. May contain adenine, cytosine, guanine, and thymine. 1
6 Carry genetic information from nucleus to ribosomes. 2
7. Are present in the nucleus of the cell. 3
8. Consist of chains of nucleotides. 3
9. What is the process called which uses the information coded in DNA to chemically link amino acids into a chain?
translation
10. The amino acid sequences of three species shown below were determined in an investigation of evolutionary
relationships.
Species A: Val His Leu Ser Pro Val Glu
Species B: Val His Leu Cys Pro Val Glu
Species C: Val His Thr Ser Pro Glu Glu
Based on these data, which two species are most closely related? Support your answer.
Species A & B because they have more amino acids in common
30
Unit Review
1. A child is born with a genetic disorder to parents who show no symptoms of the disorder. Explain the type of
information a genetic counselor might provide to these parents. In your answer, be sure to:
 explain why the child exhibits symptoms of the genetic disorder even though the parents do not
 identify one technique that can be used to detect a genetic disorder
 identify one genetic disorder

it’s a recessive trait/ parents are heterozygous/ mutation occurred/ nondisjunction occurred

karyotyping/ blood sampling/ amniocentesis

Downs Syndrome/ hemophilia/ sickle cell anemia/ PKU
2. One variety of wheat is resistant to disease. Another variety contains more nutrients of benefit to humans.
Explain how a new variety of wheat with disease resistance and high nutrient value could be developed. In your
answer, be sure to:
 identify one technique that could be used to combine disease resistance and high nutrient value in a new
variety of wheat
 describe how this technique would be carried out to produce a wheat plant with the desired characteristics
 describe one specific difficulty (other than stating that it does not always work) in developing a new
variety using this technique

genetic engineering/ selective breeding

take gene for desirable trait and move it into another plant/ cross two plants with desirable traits

difficult to isolate one gene/ other traits may show up/ moved gene may not be expressed
3. Although human muscle cells and nerve cells have the same genetic information, they perform different
functions. Explain how this is possible.
Some genes are turned on or expressed/ cells are specialized
4. For many years, humans have used a variety of techniques that have influenced the genetic makeup of organisms.
These techniques have led to the production of new varieties of organisms that possess characteristics that are
useful to humans. Identify one technique presently being used to alter the genetic makeup of an organism, and
explain how humans can benefit from this change. Your answer must include at least:
 the name of the technique used to alter the genetic makeup
 a brief description of what is involved in this technique
 one specific example of how this technique has been used
 a statement of how humans have benefited from the production of this new variety of organism

genetic engineering/ gene manipulation/ selective breeding/ gene therapy

move segment of DNA from one organism into another

production of insulin/ human growth hormone

more insulin available for diabetics
5. Knowledge of human genes gained from research on the structure and function of human genetic material has led
to improvements in medicine and health care for humans.


state two ways this knowledge has improved medicine and health care for humans
identify one specific concern that could result from the application of this knowledge

diagnose diseases/ prevent diseases/ gene therapy/ genetic engineering to produce hormones

overpopulation/ may lead to discrimination/ may limit insurance coverage
31
Base your answer to questions 6 and 7 on the information and chart.
6. Identify one environmental factor that could cause a base sequence in DNA to be changed to a different base
sequence.
carcinogen, mutagen, chemicals. UV rays, radiation, X-rays
7. Describe how a protein would be changed if a base sequence mutates from GGA to TGA.
different amino acid sequence, Threonine instead of proline
Base your answers to questions 8 and 9 on the following passage.
The Human Genome Project
For a number of years, scientists at Cold Spring Harbor Laboratory have been attempting to map every known
human gene. By mapping, scientists mean that they are trying to find out on which of the 46 chromosomes each
gene is located and exactly where on the chromosome the gene is located. By locating the exact positions of
defective genes, scientists hope to cure diseases by replacing defective genes with normal ones, a technique known
as gene therapy. Scientists can use specific enzymes to cut out the defective genes and insert the normal genes.
They must be careful to use the enzyme that will splice out only the target gene, since different enzymes will cut
DNA at different locations.
While the human genome project should eventually improve the health of humans, many people are skeptical and
apprehensive, believing that gene therapy would be working against nature and would have religious, moral, legal,
and ethical implications.
8. Using one specific example, explain why the human genome project is considered important.
to cure diseases/ you can replace defective genes with normal ones
9. Explain why scientists must use only certain enzymes when inserting or removing a defective gene from a cell.
different enzymes will cut DNA at different locations
10. Give three examples of how the technology of genetic engineering allows humans to alter the genetic makeup of
organisms.
produce insulin/ clone sheep/ produce disease-resistant plants
32