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Unit 3
Genetics
1
Gregor Mendel


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Austrian Monk 1800’s
Observed some traits
disappeared in one
generation, only to reappear
in the next
Hypothesis: Some traits
are stronger than others.
Experimental Design


Needed something which
could be easily
manipulated.
Something with a variety of
visible characteristics.
2
Gregor Mendel
http://www.youtube.com/watch?v=MWkFxWXHTnw&playnext_from=TL&videos=c5-UkltqOnI
3
Mendel Chose the Garden Pea

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Quick generation time
Does not require much
space
Peas undergo selffertilization
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Pollinate themselves
Could also pry open petal to
make specific crosses
between plants.
Many visible traits:
Flower color
Seed color
Seed shape
Plant height
4
Mendel’s Experimental Design

Parental Generation
 For each trait studied he wanted a true breeding plant to begin
the experiment.



True BreedingGeneration after generation breeds true to a single
visible trait.
Experiment
 Cross two different parental generation plants (each One displays
One version of the trait)
 Parental white flower x purple flower cross them and plant the
seeds to get the first generation
Recorded The Numbers & Traits Of All Generations
 F1—First Filial (Latin for son/daughter) all purple 100%.
 Allowed F1 to self-fertilize.
 Counted the F2 for number of each trait.
 75% purple, 25% white.
5
Mendel’s Experiment

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In this cross Mendel
used Parental Tall
(TT) and short (tt)
All the F1 generation
plants were Tall
These F1 plants self
fertilized to produce
the F2 plants
The F2 generation
had 75% Tall and
25% short
6
Terminology

Allele – different versions of a trait




Example: Pea plants have white or purple
flowers.
White and Purple are different alleles for the trait
of flower color
Genotype – the sum of all alleles in an
individual
Phenotype – the physical representation of
the genotype (what the individual looks like)
7
Terminology

Homozygous – the two alleles for a given trait
are the same in an individual


In our flowers PP homozygous purple or pp
homozygous for white
Heterozygous – the two alleles for a given
train are different in an individual

In our flowers Pp heterozygous for flower color
purple
8
Terminology

Dominant – the allele expressed as the phenotype
in a heterozygote individual



Denoted by using the upper case of the letter for the trait
The Pp plant will be Purple
Recessive – the allele not expressed in a
heterozygote

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Denoted by using the lower case of the letter for the trait
The lower case p denotes the recessive white allele
9
Terminology Examples

Pea plants have either white or purple flowers

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Purple is dominant to white
A purple plant can be homozygous

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P – purple p – white
PP for flower color
OR heterozygous

Pp for flower color
10
Terminology Examples

A white flower plant must be homozygous
(pp) for flower color

The recessive trait will only be seen as the
phenotype when the individual has two copies of
that recessive allele
11
Meiosis

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Specialized cell division for sexual
reproduction
Results in the production of haploid gametes
Each gamete will have a single copy of each
gene
12
Determining Possible Gametes

Start with the genotype of each parent
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Parent 1 PP; Parent 2 pp
Each of parent 1’s gametes will contain 1 P
Each of parent 2’s gametes will contain 1 p
Parent 1 Gametes
P
P
Parent 2 Gametes
p
p
13
Determining Possible Gametes

The parental generation cross between these
two plants PP x pp will result in all progeny
being Pp

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They will be phenotypically purple
They are all heterozygous for flower color
Pp
14
Determining Possible Gametes


These first generation (F1) plants will then
self fertilize to create the second (F2)
generation
Start with the genotype of each parent

P

p

Parent 1 Pp; Parent 2 Pp
Half of parent 1’s gametes will contain P the other
half will contain p
Half of parent 2’s gametes will contain P the other
half will contain p
P
p
15
Punnett Squares

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Graphical representation of each parents
gametes
Allows for prediction of possible progeny for a
given set of parents
Determine the gametes of the parents
Arrange the gametes on a grid to predict the
genotypes and phenotypes of the next
generation
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Punnett Squares

Arrange the gametes of
one parent across the
top of the grid
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Punnett Squares
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Arrange the gametes of
the other parent down
the side of the grid
18
Punnett Squares

Fill in the boxes of the
parental gametes from
the top down through
the boxes
19
Punnett Squares

Fill in the boxes of the
parental gametes from
the side across
through the boxes
20
Punnett Squares

The interior of the
boxes now has the
possible genotypes of
the progeny from this
parental cross of Pp
x Pp
21
Punnett Squares

Genotypically
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25% Homozygous Purple
50% Heterozygous Purple
25% Homozygous White
Homozygous
Purple
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Phenotypically
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Heterozygous
Heterozygous
Purple
Homozygous
75% Purple
25% White
Purple
White
22
Test Cross

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Used to determine an unknown individual’s
genotype.
Cross unknown with a known homozygous.
Which homozygous? A dominant or
recessive? … see next slide for answer
23
Test Cross
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*Homozygous Recessive!
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Allows you to deduce the genotype by offspring
produced.
If any offspring are recessive then the unknown
must have been Heterozygous.
If all offspring are dominant the unknown purple
plant is homozygous dominant
24
Test Cross

You found a purple pea
plant …

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Is it homozygous or
heterozygous for flower
color?
Test cross is with a
white pea plant
If it is homozygous
purple
If it is heterozygous
purple
25
Two Factor Crosses
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A two factor cross looks at two different traits
at the same time for example seed shape
(round or wrinkled) AND plant height (tall or
short)
Set up is the same for the punnett square
Keep the alleles for each trait together in the
boxes!!
26
Two Factor Crosses
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Peas can be either round (R) or wrinkled
(r)
Plants can be tall (T) or short (t)
A plant can be homozygous for either or both
of these traits (RRTT, RRtt, rrTT,rrtt)
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round tall, round short, wrinkled tall, wrinkled short
A plant can be heterozygous for either or both
of these traits (RrTt, RRTt, RrTT,)

round tall for all of them
27
Two Factor Crosses
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We’ll cross two double heterozygous plants
(RrTt x RrTt)
Each gamete will have one allele of each trait
The gametes produced by each of these
individuals are as follows
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RT
Rt
rT
rt
28
Two Factor Crosses
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Arrange the gametes of one parent across the top of the grid
Arrange the gametes of the other parent down the side of the grid
29
Two Factor Crosses

Fill in the boxes with the gametes from the top down through the
boxes
30
Two Factor Crosses

Fill in the boxes with the gametes from the side across through
the boxes
31
Two Factor Crosses

Determine all of the possible phenotypes and
genotypes for the progeny
32
Two Factor Crosses
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Phenotypes
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Round and Tall
Round and Short
Wrinkled and Tall
Wrinkled and Short
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Genotypes
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Round / Tall
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Round / Short
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RRtt
Rrtt
Wrinkled / Tall
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RRTt
RrTt
RRTT
rrTT
rrTt
Wrinkled / Short
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rrtt
33
Standard Dominance
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In the heterozygote, the dominant allele is
expressed as the phenotype
Purple flower color is dominant to white

A Pp plant will be purple
34
Incomplete Dominance
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The heterozygote individual has a phenotype
in between the two phenotypes
Flower color of roses
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R red
r white
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RR = red
Rr = pink
Rr = white
35
Incomplete Dominance
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RR x rr
F1 all Rr = pink
Cross the F1s Rr x Rr
F2
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25% Red
50% Pink
25% White
36
CoDominance
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Neither allele is
dominant
Heterozygote
expresses both alleles
equally
In flowers a red and
white flower
ABO blood groups
37
CoDominance
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There are 3 alleles for blood type
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IA – A
IB – B
i–O
Each person has 2 alleles
The combination of the 2 alleles
determines the individual’s blood type
38
CoDominance
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Phenotype A has 2 possible genotypes
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Phenotype B has 2 possible genotypes
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IA IA
IA i
IB IB
IB i
Phenotype O has only 1 possible
genotype

ii
39
CoDominance
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If mom is IA IA and dad is IB IB all of their
children will be type IA IB
If mom is IA i and dad is IB i what are
the possible blood types of their
children?
40
CoDominance
Set up a punnett square as normal with
mom’s alleles across the top and dad’s
alleles down the side
41
CoDominance
Fill in the boxes as before
42
CoDominance
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Genotypes
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AB
Bi
Ai
ii
Phenotypes
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AB
B
A
O
43
CoDominance
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Questions to ponder
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Female blood type A, Male blood type AB
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What blood type(s) is/are not possible for their children?
Male blood type O
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What blood type(s) is/are not possible for their children?
44
CoDominance
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Questions to ponder
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Female blood type A, Male blood type AB
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What blood type(s) is/are not possible for their children?
… type O is not possible
Because Dad is AB he does not have a copy of the
recessive i to pass down to any of his children
Male blood type O
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
What blood type(s) is/are not possible for their children?
… type AB is not possible
To have blood type O Dad must be homozygous ii so all
of his children will get one copy of i from him thus AB
blood type is not possible
45
Epistasis
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Two genes are required to produce a
particular phenotype
We will study coat colors of Labrador
retrievers
46
Epistasis
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The interaction of 2
genes determine coat
color
Pigment
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B = Black, b = Brown
Deposition
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E = yes, e = no
whether the pigment can
be integrated into the fur
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Epistasis
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Black Labs
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Must have at least one B and one E
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Brown or chocolate Labs
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Must have 2 bb and at least one E
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BBEE, BBEe, BbEE, BbEe
bbEE, bbEe
Yellow Labs
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Are homozygous recessive for deposition ee
They cannot put the pigment into their fur
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BBee, Bbee, bbee
48
Examples
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What are the genotypes of the Brown and
Black parents who only produce Black and
Yellow puppies?
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First determine what you know about the
genetics of the parent dogs and their puppies
49
Examples
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You know the black lab parent has at least
one B allele and one E allele B_E_
You know the brown parent must be
homozygous for brown pigment because it is
recessive bb and that the pigment is
deposited so at least one E so it is bbE_
50
Examples
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The black puppies have at least one B and
one E
The yellow puppies must be homozygous
recessive for deposition (ee) but you don’t
know what their pigment alleles are
51
Examples
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Parents:
Puppies:

B _ E _ x bbE _
B _E _ and _ _ ee
Remember the question is what are the
genotypes of the parents …
52
Examples
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Parents:
Puppies:
B _ E _ x bbE _
B _E _ and _ _ ee
To have any yellow puppies, each parent
must have a recessive (e) allele to give
To not have any brown puppies, all puppies
must be receiving a dominant B
53
Examples
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Parents:
Puppies:
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B B E e x bbEe
B _E _ and _ _ ee
The red indicates the alleles you figured out!
The black dog parent must be homozygous
because there were no brown puppies
When we set up the punnett square we see
we’re right …
54
Examples
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Parents:

B B E e x bbEe
Gametes: BE, Be and bE, be
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Examples
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Parents:

B B E e x bbEe
Gametes: BE, Be and bE, be
Black
Black
Black
Yellow
56
Examples

What are the genotypes of the Brown and
Yellow parents who only produce Black and
Yellow puppies?

First determine what you know about the
genetics of the parent dogs and their puppies
57
Examples


You know the yellow lab parent must be
homozygous recessive for deposition (ee)
You know the brown parent must be
homozygous for brown pigment because it is
recessive bb and that the pigment is
deposited so at least one E (bbE_)
58
Examples


The black puppies have at least one B and
one E
The yellow puppies must be homozygous
recessive for deposition (ee) but you don’t
know what their pigment alleles are
59
Examples


Parents:
Puppies:

_ _ ee x bbE _
B _E _ and _ _ ee
Remember the question is what are the
genotypes of the parents …
60
Examples


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Parents:
Puppies:
_ _ ee x bbE _
B _E _ and _ _ ee
To have any yellow puppies, each parent
must have a recessive (e) allele to give
To not have any brown puppies, all puppies
must be receiving a dominant B
61
Examples


Parents:
Puppies:

B B ee x bbEe
B _E _ and _ _ ee
The red indicates the alleles you figured out!
When we set up the punnett square we see
we’re right …
62
Examples

Parents:

B B e e x bbEe
Gametes: BE, Be and bE, be
63
Examples

Parents:

B B e e x bbEe
Gametes: BE, Be and bE, be
Black
Yellow
Black
Yellow
64
Environmental Factors Influence Gene
Expression

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Gene products are proteins.
Proteins are affected by temp, PH, salt
concentration, etc.
Some gene products appear differently in different
environments.
65
Environmental Gene Interaction
Winter
Coat
Summer Coat

Artic Fox

Protein production (Pigment) temperature regulated.
WarmProduce pigmentBrown.
ColdNo pigmentWhite.


66
Environmental Gene Interaction
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Siamese Cat
Cooler areas produce
pigment.
Ears/nose &
extremities.
If cat gets obese the
layer of fat will insulate
the skin from the body
heat, entire cat will now
be pigmented.
67
Genetic Issues

Recessive does not denote good …
Dominant does not denote good

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Perfect VisionRecessive
Huntington’s diseaseDominant
Meiosis can make errors in gamete formation
68
Meiotic Error Non-Disjunction


The failures of
chromosomes to
separate during
Anaphase I or II.
Results in one gamete
with too many
chromosomes & one
gamete with too few.
69
Non-Disjunction Examples
Down’s Syndrome: Trisomy 21

Individual has three copies of chromosome 21.

More common in older mothers.

Males produce sperm throughout their physically mature lifetime

Females are born with all their eggs.

Eggs are stalled at the end of meiosis I.

Upon physical maturity, 1-2 eggs per month complete meiosis, two are ovulatedOlder
mothers….older eggs. Any damage accumulates over the life-time and can result in non
disjunction in meiosis 2.

Nondisjunction In Sex Chromosomes
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Can occur in both male and females.
XXX Sterile female
XXY Sterile male
XYY Fertile male
X
Sterile female
Y
Non viable
70
Dominant Diseases

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Huntington’s disease is a dominant
progressive neurological disorder resulting in
death
How many alleles are required for
Huntington’s disease?
How many alleles are required for a
recessive disorder?
71
Dominant Diseases


Huntington’s disease is a dominant progressive
neurological disorder resulting in death
How many alleles are required for Huntington’s
disease?


It’s dominant so you only need one allele to have the
disorder
How many alleles are required for a recessive
disorder?

Recessive traits are only expressed if the individual is
homozygous so you need 2 alleles, one from each parent
72
Disease Terminology

Autosomal
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Sex Linked

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Trait is on one of the 22 pairs of non-sex chromosomes
These traits are inherited equally among males and
females
Trait is on either the X or Y chromosome
These traits are inherited differently based on the person’s
gender
Carrier


individual has one copy of the recessive allele but is not
afflicted by the disease
There are no carriers for dominant traits
73
Sex Linked Traits

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These traits are encoded on the X or Y chromosomes
The gender of the individual is linked to the expression of these
traits
Sex Chromosomes X, Y
 XX=Female
 XY=Male.
Male sperm carry either X or Y determines gender of offspring.
Female eggs only carry an X for sex chromosome.
Since female have two X chromosomes, they follow standard
dominance patterns for genes carried on X chromosome.
 Females cannot inherit genes on the Y chromosome because
they don’t have a Y chromosome
Males only have one X chromosome.
Any genes on their one chromosome are automatically
expressed.
74
Sex Linked Traits
http://www.youtube.com/watch?v=H1HaR47Dqfw&playnext_from=TL&videos=wNlPLX5yjQE
75
Pedigrees
Males
Normal
Carrier
Afflicted
Females
76
Pedigrees
Horizontal Line
Represents Marriage
Carrier Dad
Carrier Mom
Vertical Line Down
to their children
Afflicted
Daughter
Normal
Son
Carrier
son
77
Pedigree Example
This is an autosomal recessive disorder
?
What is the father’s most probable genotype?
78
Pedigree Example
This is an autosomal recessive disorder
Dd
DD
DD
DD
Dd
Dd
What is the father’s most probable genotype?
Mom is a carrier, they have 2 normal kids and two
Carriers … from this information dad is most probably
normal
79
Pedigree Example
This is a sex linked recessive disorder
?
What is the father’s most probable genotype?
80
Pedigree Example
This is a sex linked recessive disorder
XcX
XcY
XcY
XcXc
XY
XcX
What is the father’s most probable genotype?
A daughter is afflicted. Since this is sex linked and recessive
We know that dad must be afflicted because the daughter received
a recessive X from each of her parents.
81
How do our genes make us who we are?


Genes are the construction plans for proteins
DNA transcribed into RNA (Single stranded)


Called mRNA—Messenger RNA
Ribosomes can only read mRNA
82
Transcription
http://www.youtube.com/watch?v=vJSmZ3DsntU&playnext_from=TL&videos=g_7DQONcg2I
83
Transcription

RNA Polymerase





Enzyme which reads DNA & makes the mRNA copy.
mRNA copy made by complimentary base pairing to the
DNA
Binds to promoter at the beginning of each gene
Read DNA one base at a time
Transcription



RNA Polymerase binds to promoter
Reads one base at a time synthesizing single stranded
mRNA from the DNA template
mRNA transported to the cytoplasm through the nuclear
pores
84
Translation
http://www.youtube.com/watch?v=1NkLqjQkGHU&playnext_from=TL&videos=aJ9DK01wG2U
85
The Genetic Code

DNA 4 bases


RNA 4 bases


Adenine, Thymine, Guanine, Cytosine
Adenine, Uracil, Guanine, Cytosine
20 Amino Acids (A.A,).





How to specify each individual A.A.
Use a distinct, unique set of 3 bases.
Called a codon.
Each A.A. is coded for by at least one codon.
Special codons.


Start/F metheonine: AUG
Stop: UAA, AAG, UGA
86
The Codon Table
87
Find AUG
The first letter of the codon is A … So you know you start with A as the
First Position
88
Find AUG
The second letter of the codon is U … So you know you’ve narrowed it down
To these 4 with the Second Position
89
Find AUG
The third letter of the codon is G … So you now you know it is
Met (methionine) or start if it’s the first codon of the sequence
90
Translation

Protein synthesis



Ribosomes read the mRNA.
Assemble A.A. in order by
reading one codon at a
time.
How do the A.A. get to the
ribosomes?




tRNA—Transfer RNA
Molecules which bring A.A.
to ribosomes.
Has Anti codon at one end
to temporarily base pair to
mRNA.
Has corresponding A.A. at
other end
91
Translation Steps

Initiation


Small ribosome subunit
binds to mRNA at start
codon AUG.
Large subunit then binds
to complex.
92
Translation Steps

Elongation





Ribosome moves down mRNA one codon at a timeadding one amino
acid at a time.
tRNA comes in and binds by complimentary base pairing.
Peptide bond is formed between the new amino acid and the peptide
chain
Continues down mRNA until stop codon reached.
Transfer RNA only carries specific A.A.
93
Translation Steps

Termination


Stop codon; signal the two ribosome subunits to
break apart
Completed protein released
94
Genetic Control




Regulation Of Gene Expression
Differentiation requires that some of our
genes be turned on or off in specific cells.
Remember, each and every cell has Full set
of DNA.
Cells express different genes according to
their functions.
95
Genetic Control

Repressors




Prevents transcription.
Bind over Promoter.
Blocks RNA ploymerase from binding.
Activators



Increase rate of transcription.
Bind upstream of Promoter.
Assist RNA polymerase in binding to DNA.
96
DNA / RNA and Genetic Control
http://www.youtube.com/watch?v=BGtNZwd3brg&playnext_from=TL&videos=1Ihv5jV0kjs
97