Pedigree Studies

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12 Δεκ 2012 (πριν από 4 χρόνια και 6 μήνες)

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1

Name: _________________________________







Per. ______

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

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



nondisj
unction

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



mal
ignant



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

mRNA

o

tRNA

o

rRNA

codon

protein synthesis

o

translation


2

Part A: Fundamentals of Genetics

Gregor Mendel: Genetic Pioneer


genetics = study of
genes


Gregor Mendel
-

Austrian Mon
k

-

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 =
P
1

generation


first generation offspring =
F
1

generation


second generation offspring =
F
2
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 al
lelic 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


3

genotype: 2Bb: 2bb


phenotype: 2black: 2 white

genotype: 2Rr: 2Rr

phenotype: 4 round

genotype: 4Ii

phenotype: 4 inflated

Cross with short plant to see
if you get a short offspring

key: B= black

b= white

key: I= inflated


i= constricted

key: T= tall

t= short

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


Punnett
Squares and Probability



show possible results of crosses


X


X








X







Y

Practice:

1. cross a homozygous tall pea plant with a short pea plant.

Key:

T=
tall


t= short

2. cross the hybrid F1’s from the above cross.





3. cross a heterozygous axial
-
flowered pea plant with a terminal flowered pea plant.




4. cross a homozygous, round
-
seeded pea plant with a heterozygous round
-
seeded pea plant.




5. black coat color is dominant over white coat color in guinea pigs. Cross a
heterozygous black guinea pig with a

white guinea pig.


6. cross a constricted pod pea plant with a homozygous inflated pod pea plant.



7. how would you find out if a tall pea plant is
homozygous or heterozygous?



these are all monohybrid crosses


X
X

X
X

X
Y

X
Y


T

T

t

Tt

Tt

t

Tt

Tt


T

t

T

TT

Tt

t

Tt

tt


A

a

a

Aa

aa

a

Aa

aa


R

R

R

RR

RR

r

Rr

Rr


B

b

b

Bb

bb

b

Bb

bb


I

I

i

Ii

Ii

i

Ii

Ii


T

t

t

Tt

tt

t

Tt

tt

male

gametes

female

gametes

genotype: 1TT: 2Tt: 1tt

phenotype: 3 tall: 1 short

genotype: 4Tt

phenotype: 4 tall

genotype: 2Aa: 2aa



phenotype: 2 axial: 2 terminal

key: A= axial


a= terminal

key: R= round

r= wrinkled


4

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 re
ason 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
organ
ism’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 de
scribes 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%


5

R




R

W




W

R




r

R


r

w


w

R


r

RR



RR


Rr



Rr

Ww



Ww


Ww



Ww

RR



Rr


Rr



rr

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.






F1






genotypes:
2RR: 2Rr





phenotypes:
2red: 2 pink




6. Cross a white andalusian chicken with a black andalusian chicken.






F1






genotypes:
4 Ww





phenotypes:
4 blue






7. Cross two roan short
-
horned cattle.






F1






genotypes:
1RR: 2Rr: 1rr




phenotypes:
1 red: 2 roan: 1 white





6

G

g

g

g

T

T

T

t

T

t

t

t

R

r

r

r

r

r

r

r

B

B

B

B

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 t
he recessive color.








G G







Genotype= 2 GG: 2 Gg : 0 gg







G


g


Phenotype= 4 Green pea plants: 0 yellow pea plants


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



3. A tall plant (Tt) is crossed with a short plant (tt).





genotype= 2Tt: 2tt




phenotype= 2 tall: 2 short



4. A red flower (Rr) is crossed with a white flower (rr).





genotype= 2 Rr: 2 rr




phenotype= 2 red: 2 white


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






GG



GG



Gg



Gg

Gg

gg

Gg

gg

TT

TT

Tt

Tt

Tt

tt

Tt

tt

Rr

rr

Rr

rr

rr

rr

rr

rr

BB

BB

BB

BB


7

B

b

B




B

W

w

W




w

R

r

R




r

T


t

T




T

w


w

W




w

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.

A homozygous dominant brown mouse is crossed with a heterozygous brown mouse (tan is the recessive
color).


BB

BB

Bb

Bb



2.

Two heterozygous white (brown fur is recessive) rabbits are crossed.



WW

Ww

Ww

ww


3.

Two heterozygous red flowers (white flowers are recessive) are crossed.



RR

Rr

Rr

rr



4.

A homozygous tall plant is crossed with a heterozygous tall plant (short is the recessive size).



TT

TT

Tt

Tt


5.

A
heterozygous white rabbit is crossed with a homozygous black rabbit.



Ww

ww

Ww

ww

genotype= 2 BB: 2 Bb

phenotype= 4 brown

genotype= 1WW: 2Ww: 1ww

phenotype= 3 white: 1 brown

genotype= 1RR: 2Rr: 1rr

phenotype= 3 red: 1 white

genotype= 2 TT: 2Tt

phenotype= 4 tall

genotype= 2 Ww: 2 ww

phenotype= 2 white: 2 black


8

X


X

X


Y

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:








Are Chromosomes the Same in Male and Female?

Chromosomes in Female and Male.



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. I
n 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

XX

XX

XY

XY

X

X

X

Y


9

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 the
Y
.

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
: ch
ange 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
-
disju
nction
: 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
mak
ing a codon


10

I
A




i


I
B




i


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
deterioration
from 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:
I
A
; I
B
; ii



each person’s blood contains
2
of these alleles



I
A
, I
B

are
codominant
(both expressed when together) and are both dominant to the

i



A type =
I
A

I
A
; I
A
i



B type =
I
B
; I
B

i



AB type =
I
A

I
B



O type =
ii



Which cross could result in all four blood types in the offspring? Show results of cross below:



I
A

I
B


I
B

i

I
A
i

ii






11

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 20
-
2). 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 generati
on 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 oldes
t 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. Generat
ion 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


12

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 h
omozygous 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 pedigre
e, there are two rules you must
follow.

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 th
e persons in a pedigree. Figure 20
-
4
shows the genes predicted by using these two
rules.


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
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









13

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.

(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%

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 attach
ed 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.




Examine the pedigree:







9. (a) Wh
ich Punnett square, A, B, or C, would
best fit this family?
B

(b) Explain.
offspring with ee

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.






14

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

so you received this gene
from one parent

and this gene from the
ot
her parent

A


I
A

I
A

or


I
A
i

I
A


I
A

I
A

i

B


I
B

I
B

or


I
B
i


I
B

I
B

I
B

i

AB

I
A

I
B

I
A

I
B

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

C
an receive blood from type(s)…

A

A, O

B

B, O

AB

A, B, AB, O

O

O


Universal Donor =
O


Universal Recipient=
AB




15

I
A




I
A


I
B




I
B


I
A




I
A


i




i


I
B




I
B


I
A




I
B


I
B




i


I
A




I
B


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?








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?














3)
Could a man with type B blood and a woman with type AB produce a child with type O blood?

















I
A

I
B

I
A

I
B

I
A

I
B

I
A

I
B

I
A

i

I
A

i

I
A

i

I
A

i

I
A

I
B

I
A

I
B

I
B

I
B

I
B

I
B

I
A

I
B

I
A

i

I
B

I
B

I
B

i


16



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



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

o

representing sex
-
linked traits:



X
C

=
normal
, X
c

=
colorblind



X
H

=
normal
, X
h

=
hemophiliac



X
= normal,

X

= disease trait

Colorblindness Example:



o

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

normal female

normal female

colorblind female

normal male

colorblind male


17

X


X

XX

XX

XY

XY

X


X

XX

XX

XY

XY

Figure 1: Inheritance of Hemophilia



Figure 2: Inheritance of Hemophilia

“Carrier” Mother and Father Without 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
hemophilia
c
.

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%


b) What percent of their sons?




100%


c) What percent of their daughters?




0%





2) Neither Marie or Frank is colorblind but their son Raymond is
colorblind.


How is this possible?

Marie is a carrier









X




Y



X




Y


18



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

o

XYY
: tall aggressive males, criminals?

o

Down Syndrome
: Trisomy 21



extra chromosome 21



mild to severe retardation, facial features, muscle weakness, heart defects, short stature


Monosomy X




Tri
somy X




Trisomy 21




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


19


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 t
he 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+ Rh
-

Rh+ (positive is dominant)


Mm


darker skin


Cc


normal vision

Which parent do
nated 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


20

Determine the genotypes and corresponding phenotypes of the people whose chromos
omes 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
+




Rh
+


f.

mm



light skin


g.

CC



normal vision










Genotypes




Phenotypes

2.




a.


Ff



freckled




b.

TT



can taste




c.

bb



blond hair




d.

hh



no disease



e.

Rh
+
Rh
-




Rh
+




f.

mm



light skin




g.

Cc




normal vision




Genotypes




Phenotypes

3.




a.

ff



unfreckled




b.

tt



can’t taste




c.

BB



brown hair




d.

HH



has disease



e.

Rh
-
Rh
-




Rh
-




f.

MM



dark skin




g.

cc




colorblind



21

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,


6.
the same


would have more freckles?

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

9.
yes
(TT)


have taste ability?

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
cur
e 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 s
ticky ends form

recombinant

form


out of DNA from 2
organisms

o

cloning vectors: carrier used to clone a gene and transfer it to another organism

o

plasmid:
ring
of DNA in a bacterium




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


22

2. DNA Technology Techniques



DNA Fing
erprints:

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 Fing
er 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 th
e 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 nitrogen
-
poor 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

controlled
by gene expression


23



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

sarcomas
:
bone
and muscl
e tissue

o

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 techniq
ues 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 plan
ts 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 go
lden 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 poll
inated 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 f
rom 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


24

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, the
y exhibit normal Himalayan hair pattern. However, when their offspring are exposed to
low temperatures (10
o
C), they have black fur covering their entire bodies. This illustrates:


1. that mutations are caused by heat radiation


2. that the traits of an org
anism 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. chr
omosomes


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. artifi
cial 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

21
st

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 selecti
on

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 w
ould 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



25

Part F

Nucleic Acid and Protein
Synthesi
s

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


a
ll

t
eachers





G
----
C


g
o

c
razy

The Roles of DNA and RNA

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


26


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


-
ribos
e 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)



27

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



3. thymine


4. uracil

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 a
t 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 m
olecule 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 nitrogeno
us 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. po
lyploidy

14.

Which nuclear process is represented below?



1. recombination

2. fertilization


3. replication


4. mutation


28

Protein Synthesis

Protein synthesis is a complex process. You will trace the steps that are involved in the protein synthesis of a p
art
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 n
ucleus. 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



ACA
-

ATA
-

TAG
-

CTT
-

TTG
-

ACG
-

GGG
-

AAC
-

CCC
-

ATT




1


2 3 4 5


6


7


8


9 10


TRANSCRIPTION:

1


2 3 4 5



6


7


8


9 10


mRNA:
(Codon)
UGU



UAU


-

AUC


-

GAA


AAC
-

UGC


CCC
-

UUG

-

GGG

-

UAA


B.
After
transcription

(
in nucleus
), mRNA
attaches

to a
ribosome

where
translation

(in cytoplasm)
takes place. Each
codon
o
f 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 t
he 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

mRNA codon

tRNA codon

Amino Acid

AAA

UUU

AAA

Phenylalanine

GTC

CAG

GUC

glutamine

ACT

UGA

ACU

STOP

UAC

AUG

UAC

Methionine (START)

mRNA

1.
UGU

2.
UAU

3.
AUC

4.
GAA

5.
AAC

6.
UGC

7.
CCC

8
UUG

9.
GGG

10.
UAA


2
9

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
UA
A
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




















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

DNA


mRNA


ribosome

translation

RNA code


amino acids

cytoplasm

protein modified

polypeptide chains folded to into proteins




DNA

mRNA

tRNA

mRNA

ribosome


30

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. hy
drolysis


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 R
NA 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



31

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 ev
en 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/ amnioc
entesis



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 de
sired 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 o
f 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 organis
m



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


32

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 chang
ed 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 techniqu
e 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 loc
ations
.

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 defectiv
e 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